Calculus Third Edition Gujarat Technological University 2017 9789352607310, 9352607317

Table of contents :
Title
Contents
Unit 1 Infinite Sequences and Series
1 Sequences and Series
2 Taylor’s and Maclaurin’s Series
Unit 2 Curve Sketching
3 Curve Sketching
Unit 3 Indeterminate Forms & Improper Integrals
4 Indeterminate Forms
5 Improper Integrals
Unit 4 Applications of Integration
6 Applications of Integration
Unit 5 Partial Derivatives
7 Partial Derivatives
8 Applications of Partial Derivatives
Unit 6 Multiple Integrals
9 Multiple Integrals
Appendix 1 Differential Formulae
Appendix 2 Integral Formulae
3 Standard Curves
Question Papers
Index

Citation preview

Calculus Third Edition Gujarat Technological University 2017

About the Authors Ravish R Singh is presently Academic Advisor at Thakur Educational Trust, Mumbai. He obtained a BE degree from University of Mumbai in 1991, an MTech degree from IIT Bombay in 2001, and a PhD degree from Faculty of Technology, University of Mumbai, in 2013. He has published several books with McGraw Hill Education (India) on varied subjects like Engineering Mathematics, Applied Mathematics, Electrical Networks, Network Analysis and Synthesis, Basic, Electrical Engineering, Basic Electrical and Electronics Engineering, etc., for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Dr A P J Abdul Kalam Technical University. Dr Singh is a member of IEEE, ISTE, and IETE, and has published research papers in national and international journals. His fields of interest include Circuits, Signals and Systems, and Engineering Mathematics. Mukul Bhatt is presently Assistant Professor, Department of Humanities and Sciences, at Thakur College of Engineering and Technology, Mumbai. She obtained her MSc (Mathematics) degree from H N B Garhwal University in 1992, and a PhD degree from Faculty of Science, PAHER University, Udaipur, Rajasthan in 2017. She has published several books with McGraw Hill Education (India) Private Limited on Engineering Mathematics and Applied Mathematics for all-India curricula as well as regional curricula of some universities like Gujarat Technological University, Mumbai University, Pune University, Jawaharlal Nehru Technological University, Anna University, Uttarakhand Technical University, and Uttar Pradesh Technical University. Dr. Bhatt has twenty six years of teaching experience at various levels in engineering colleges and her fields of interest include Integral Calculus, Complex Analysis, and Operation Research. She is a member of ISTE.

Calculus Third Edition Gujarat Technological University 2017

Ravish R Singh Academic Advisor Thakur Educational Trust Mumbai, Maharashtra Mukul Bhatt Assistant Professor Department of Humanities and Sciences Thakur College of Engineering and Technology Mumbai, Maharashtra

McGraw Hill Education (India) Private Limited Chennai McGraw Hill Education Offices Chennai new York St Louis San Francisco auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai 600 116 Calculus, 3e, GTU–2017 Copyright © 2018, 2017, 2015 by McGraw Hill Education (India) Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listing (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited. ISBN 13: 978-93-5260-731-0 ISBN 10: 93-5260-731-7 Managing Director: Kaushik Bellani Director—Science & Engineering Portfolio: Vibha Mahajan Senior Portfolio Manager: Hemant Jha Portfolio Manager: Navneet Kumar Senior Manager—Content Development: Shalini Jha Content Developer: Sahil Thorpe Production Head: Satinder S Baveja Copy Editor: Taranpreet Kaur Assistant Manager—Production: Anuj K Shriwastava General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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Visit us at: www.mheducation.co.in

Dedicated to My Mother Late Shrimati Premsheela Singh (Ravish R Singh)

to My Sister Vijay Luxmi Hemdan (Mukul Bhatt)

Contents Preface Roadmap to the syllabus

Unit i

infinite SeqUenCeS And SerieS

1. Sequences and Series 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14

xi xv

1.1–115

Introduction 1.1 Sequence 1.1 Infinite Series 1.8 The nth Term Test for Divergence 1.9 Geometric Series 1.10 Standard Limits 1.16 Comparison Test 1.17 D’Alembert’s Ratio Test 1.36 Cauchy’s Root Test 1.63 Cauchy’s Integral Test 1.71 Alternating Series 1.77 Absolute Convergence of a Series 1.84 Uniform Convergence of a Series 1.91 Power Series 1.94 Points to Remember 1.109 Multiple Choice Questions 1.111

2. taylor's and Maclaurin's Series

2.1–70

2.1 Introduction 2.1 2.2 Taylor’s Series 2.1 2.3 Maclaurin’s Series 2.27 Points to Remember 2.67 Multiple Choice Questions 2.68

Unit 2

CUrVe SKetCHinG

3. Curve Sketching 3.1 Introduction 3.1 3.2 Monotonic Functions 3.1 3.3 Concavity, Convexity and Points of Inflection of a Curve 3.1

3.1–46

viii

Contents

3.4 3.5 3.6 3.7

Unit 3

Maxima and Minima 3.2 Tracing of Cartesian Curves 3.4 Tracing of Parametric Curves 3.24 Tracing of Polar Curves 3.31 Multiple Choice Questions 3.43

indeterMinAte forMS & iMproper inteGrAlS

4. indeterminate forms 4.1 Introduction 4.1 4.2 L’Hospital’s Rule

4.1–81

4.1

0 Form 4.2 0 • Type 2: Form 4.24 • Type 3 : 0 × • Form 4.31 Type 4 : • − • Form 4.38 Type 5 : 1•, • 0, 0 0 Form 4.46 Solving Indeterminate Forms Using Expansion 4.68 Points to Remember 4.79 Multiple Choice Questions 4.79

4.3 Type 1 : 4.4 4.5 4.6 4.7 4.8

5. improper integrals 5.1 5.2 5.3 5.4 5.5 5.6

5.1–24

Introduction 5.1 Improper Integrals 5.1 Improper Integrals of the First Kind 5.1 Improper Integrals of the Second Kind 5.9 Improper Integral of the Third Kind 5.16 Convergence and Divergence of Improper Integrals 5.17 Multiple Choice Questions 5.22

Unit 4 AppliCAtionS of inteGrAtion 6. Applications of integration 6.1 6.2 6.3 6.4

Introduction 6.1 Volume by Slicing 6.1 Volume of Solid of Revolution 6.6 Volume by Cylindrical Shells 6.33 Multiple Choice Questions 6.41

6.1–43

Contents

Unit 5

pArtiAl deriVAtiVeS

7. partial derivatives 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9

7.1–169

Introduction 7.1 Functions of Two or More Variables 7.1 Limit and Continuity of Functions of Several Variables 7.2 Partial Derivatives 7.11 Higher-Order Partial Derivatives 7.11 Variables to be Treated as Constants 7.55 Total Derivatives 7.70 Implicit Differentiation 7.105 Euler’s Theorem for Homogeneous Functions 7.114 Points to Remember 7.162 Multiple Choice Questions 7.163

8. Applications of partial derivatives 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

ix

8.1–121

Introduction 8.1 Tangent Plane and Normal to a Surface 8.1 Linear Approximation or Linearization 8.10 Errors and Approximations 8.13 Maximum and Minimum Values by Second Derivative Test 8.21 Maximum and Minimum Values with Constrained Variables 8.38 Method of Lagrangian Multipliers 8.49 Taylor’s Formula For Two Variables 8.81 Jacobians 8.97 Points to Remember 8.115 Multiple Choice Questions 8.117

Unit 6 MUltiple inteGrAlS 9. Multiple integrals 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

Introduction 9.1 Double Integrals Over Rectangles 9.1 Change of Order of Integration 9.31 Double Integrals in Polar Coordinates 9.66 Change of Variables 9.77 Triple Integrals 9.109 Area as Double Integral 9.141 Volume as Triple Integral 9.169 Points to Remember 9.188 Multiple Choice Questions 9.190

9.1–194

x

Contents

Appendix 1 Differential Formulae Appendix 2 Integral Formulae Appendix 3 Standard Curves GTU Solved Question Paper – Winter 2016 GTU Solved Question Paper – Summer 2017 Index

A1.1 A2.1–A2.3 A3.3–A3.5 SQP.1–SQP.7 SQP.1–SQP.7 I.1–I.3

preface Calculus is a key area of study in any engineering course. A sound knowledge of this subject will help engineering students develop analytical skills, and thus enable them to solve numerical problems encountered in real life, as well as apply mathematical principles to physical problems, particularly in the field of engineering.

Users This book will be useful for first-year engineering students of Gujarat Technological University (GTU). It covers the complete GTU syllabus for the course on Calculus, which is common to all the engineering branches.

Objective Crisp, complete explanations of concepts seek to enable students to easily understand, and thereby build a strong foundation in Calculus and its applications. The tutorial approach followed in this text will help them develop a logical perspective to solving problems. Solved question papers will help students understand the patterns of examinations.

FeatUres Each topic has been explained from the examination point of view, wherein the theory is presented in an easy-to-understand student-friendly style. Full coverage of concepts is supported by numerous solved examples with varied complexity levels. The solutions of examples are set following a ‘tutorial’ approach, which will make it easy for students from any background to easily grasp the concepts. Exercises with answers immediately follow the solved examples enforcing a practice-based approach. We hope that the students will gain logical understanding from solved problems and then reiterate it through solving similar exercise problems themselves. The unique blend of theory and application caters to the requirements of both the students and the faculty.

HigHligHts ∑ Crisp content strictly as per the latest GTU syllabus of Calculus ∑ Excellent coverage with lucid presentation style ∑ New! Multiple Choice Questions are incorporated within each chapter as per GTU paper style ∑ Solutions of GTU examination papers from 2010 to 2017 are present appropriately within the chapters

xii

Preface

∑ Solution to latest Summer 2017 and Winter 2016 GTU question paper are placed at the end of the book ∑ Rich exam-oriented pedagogy:  180+ Enhanced New Multiple Choice Questions  475+ Solved Examples within chapters  150+ Exam oriented Solved GTU Questions tagged within chapters  1450+ Unsolved Exercises

cHapter OrganisatiOn The content spans the following nine chapters which wholly and sequentially cover each module of the syllabus. ∑ Chapter 1 explains Sequences and Series. ∑ Chapter 2 presents Taylor’s and Maclaurin’s Series. ∑ Chapter 3 discusses Curve Tracing. ∑ Chapter 4 presents Indeterminate Forms. ∑ Chapter 5 covers Improper Integrals. ∑ Chapter 6 deals with Applications of Integration. ∑ Chapter 7 presents Partial Derivatives. ∑ Chapter 8 explains Applications of Partial Derivatives. ∑ Chapter 9 discusses Multiple Integrals. ∑ Appendices 1, 2, and 3 provide Differential Formulae, Integral Formulae, and Standard Curves, respectively.

Acknowledgements Special thanks to the reviewers mentioned here for providing encouraging comments and valuable suggestions regarding improvement of manuscript. Manokamna Agrawal Prakash Kumar Patel Hiren Bhatt Bhavini Pandya Usha Bag

Silver Oak College of Engineering and Technology, Ahmedabad, Gujarat Babaria Institute of Technology, Vadodara, Gujarat Marwadi Education Foundation Group of Institutions, Rajkot, Gujarat Sardar Vallabhbhai Patel Institute of Technology, Anand, Gujarat Shree L R Tiwari College of Engineering, Maharashtra.

We would also like to thank all the staff at McGraw Hill Education (India), especially Vibha Mahajan, Shalini Jha, Sahil Thorpe, Satinder Singh Baveja, Taranpreet Kaur and Anuj Shriwastava for coordinating with us during the editorial, copyediting, and production stages of this book.

Preface

xiii

Our acknowledgements would be incomplete without a mention of the contribution of all our family members. We extend a heartfelt thanks to them for always motivating and supporting us throughout the project. Constructive suggestions for the improvement of the book will always be welcome. Ravish R Singh Mukul Bhatt

publisher’s note Remember to write to us. We look forward to receiving your feedback, comments and ideas to enhance the quality of this book.You can reach us at [email protected]. Please mention the title and authors’ name as the subject. In case you spot piracy of this book, please do let us know.

roadmap to the Syllabus This text is useful for: Calculus (2110014) as taught in Gujarat Technological University Unit 1: Infinite Sequences and Series ∑ ∑ ∑ ∑ ∑ ∑ ∑

Introduction of convergence, divergence of sequences and infinite series The nth term test for divergence; Integral test Comparison test, Ratio test, Root test Alternating series, Absolute convergence, Conditional convergence Power series and Radius of convergence Taylor’s series Maclaurin’s series Go to: CHApter 1. Sequences and Series CHApter 2. taylor's and Maclaurin's series

Unit 2: Curve Sketching ∑ ∑ ∑ ∑

Concavity Curve sketching Polar coordinates, Relation between polar and Cartesian coordinates Graphs in polar coordinates Go to: CHApter 3. Curve Sketching

Unit 3: indeterminate forms: ∑ ∑

Indeterminate form: (0/0, •/•, • ◊ 0, • - •) Indeterminate form: (00, •0, 1•)

improper integral: ∑ ∑

Improper integrals of Type-I and Type-II Convergence and divergence of improper integrals Go to: CHApter 4. indeterminate forms CHApter 5. improper integrals

xvi

Roadmap to the Syllabus

Unit 4: Applications of integration ∑ ∑ ∑ ∑

Volume Volume Volume Volume

by slicing of solids of revolution by disk method of solids of revolutions by washer method by cylindrical shell

Go to: CHApter 6. Applications of integration

Unit 5: partial derivatives ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑

Function of two variables, graphs, level curves Limit, continuity of function of several variables Partial derivatives and Clairaut's theorem Tangent plane, Normal line Linear approximation, Total differential Chain rule, Implicit differentiation Euler’s theorem for homogeneous function Maximum and minimum values by second derivative test Lagrange multipliers Taylor’s formula for two variables Go to: CHApter 7. partial derivatives CHApter 8. Applications of partial derivatives

Unit 6: Multiple integrals ∑ ∑ ∑ ∑ ∑ ∑ ∑

Double integrals over rectangles and Fubini’s theorem, Properties of double integrals Double integrals over general region Double integrals in polar coordinates Triple Integrals, Triple integrals in cylindrical coordinates Triple integrals in spherical coordinates Change of order of integration Jacobian of several variables, Change of variable in multiple integrals Go to: CHApter 9. Multiple integrals

Unit 1 Infinite Sequences and Series

CHAPTER

Sequences and Series 1.1

1

INTRODUCTION

In this chapter, we will learn about the convergence and divergence of sequence and series. There are various methods to test the convergence and divergence of an infinite series. We will study Comparison Test, D’Alembert’s ratio test, Cauchy’s root test and Cauchy’s integral test. We will also study alternating series, absolute and uniform convergence of the series and power series.

1.2

SEQUENCE

An ordered set of real numbers as u1, u2, u3, ……..un, …… is called a sequence and is denoted by {un}. If the number of terms in a sequence is infinite, it is said to be an infinite sequence, otherwise it is a finite sequence and un is called the nth term of the sequence.

1.2.1

Limit of a Sequence

A sequence {un} tends to a finite number l as n Æ • if for every Œ > 0 there exists an integer m such that, | un - l | < Œ for all n > m, i.e., lim un = l. n Æ•

1.2.2

Convergence, Divergence and Oscillation of a Sequence

(i) If the sequence {un} has a finite limit, i.e., lim un is finite, the sequence is said n Æ• to be convergent.

e.g.

Ï ¸ Ô 1 Ô {{un } = Ì ˝ Ô1 + 1 Ô Ó n˛ lim un = 1 n Æ•

Since limit is finite, the sequence is convergent.

1.2

Chapter 1

Sequences and Series

(ii) If the sequence {un} has infinite limit, i.e., lim un is infinite, the sequence is said n Æ• to be divergent. e.g.

{un} = {{22 n + 1} lim un = •

n Æ•

Since limit is infinite, the sequence is divergent. (iii) If the limit of the sequence {un} is not unique, the sequence is said to be oscillatory. e.g.

{{un} = ( -1)n +

1 2n

lim un = 1, if n is even

n Æ•

= –1, if n is odd Since limit is not unique, the sequence is oscillatory.

1.2.3

Monotonic Sequence

A sequence is said to be monotonically increasing if unn++1 ≥ un for each value of n and is monotonically decreasing if unn++1 £ un for each value of n. The sequence is called alternating sequence if the terms are alternate positive and negative. e.g. (i) 1, 2, 3, 4, … is a monotonically increasing sequence. 1 1 1 (ii) 1, , , , … is a monotonically decreasing sequence. 2 3 4 (iii) 1, –2, 3, – 4, … is an alternating sequence.

1.2.4

Bounded Sequence

A sequence {un} is said to be a bounded sequence if there exists numbers m and M such that m < un < M for all n.

Note 1: Every convergent sequence is bounded but the converse is not true. Note 2: A monotonic increasing sequence converges if it is bounded above and diverges to + • if it is not bounded above. Note 3: A monotonic decreasing sequence converges if it is bounded below and diverges to – • if it is not bounded below. Note 4: If sequence {un} and {vn} converges to l1 and l2 respectively then (i) Sequence {un + vn} converges to l1 + l2 (ii) Sequence {un . vn} converges to l1◊ l2 Ïu ¸ l (iii) Sequence Ì n ˝ converges to 1 provided l2 π 0 v l Ó n˛ 2

1.2

Example 1

ÏÔ n2 + n ¸Ô ˝. 2 ÔÓ 2 n - n Ô˛

Test the convergence of the sequence Ì Solution un =

Let

n2 + n 2n2 - n

lim un = lim

n Æ•

n2 + n

2n2 - n 1 1+ n = lim 1 n Æ• 2n 1 = 2 n Æ•

Hence, {un} is convergent.

Example 2 Test the convergence of the sequence {tanh n}. Solution Let

un = tanh n lim un = lim tanh n

n Æ•

n Æ•

sinh n n Æ• cosh n

= lim = lim

n Æ•

= liim m

en - e- n en + e- n e2 n - 1

e2 n + 1 1 1 - 2n e = lim 1 n Æ• 1 + 2n e =1 n Æ•

Hence, {un} is convergent.

Example 3 Test the convergence of the sequence {2n}.

Sequence

1.3

1.4

Chapter 1

Sequences and Series

Solution Let un= 2n lim un = lim 2 n

n Æ•

n Æ•

=•

Hence, {un} is divergent.

Example 4

{

}

Test the convergence of the sequence 2 - ( -1)n . Solution un= 2 –(–1)n

Let

lim un = lim 2 - (( 1)n

n Æ•

n Æ•

=2–1=1

,

= 2 –(–1) = 3 ,

if n is even if n is odd

Since limit is not unique, the sequence {un} is oscillatory.

Example 5

1 1 1 Show that the sequence {un} whose nth term is un = 1 + + 2 +  + n , 3 3 3 is monotonic increasing and bounded. Is it convergent?

Solution 1 1 1 + ++ n 3 32 3 1 1 1 1 = 1 + + 2 +  + n + n +1 3 3 3 3

un = 1 + un +1

unn++1 - un =

1 3

n +1

>0

Hence, {un} is monotonic increasing sequence. Also,

un = 1 +

1 1 1 + ++ n 3 32 3

1 ˆ Ê 1 Á 1 - n +1 ˜ Ë 3 ¯ = 1 13

1.2

=

Sequence

1.5

3Ê 1 ˆ 3 1 - n +1 ˜ < Á Ë 2 3 ¯ 2

3 . 2 Since {un} is monotonic increasing and bounded above, it is convergent. {un} is bounded above by

Example 6

1 2 1 + + …+ , 1! 2 ! n! n Œ N, is monotonic increasing and bounded. Is it convergent?

Show that the sequence {un} whose nth term is un = Solution 1 1 + + …+ 1! 2 ! 1 1 un +1 = + + … + 1! 2 ! 1 unn++1 - un = >0 (n + 1)! un+1 > un un =

1 n! 1 1 + nn!! (n + 1)!

Hence, {un} is a monotonic increasing sequence. Also,

1 1 1 1 + + ++ 1! 2 ! 3! n! 1 1 1 = 1+ + ++ 2 ! 3! n! 1 1 1 < 1+ + 2 ++ n 2 2 2 1 ˆ Ê 1 Á 1 - n +1 ˜ Ë 2 ¯ < 1 12 1 ˆ Ê < 2 Á 1 - n +1 ˜ Ë 2 ¯

un =

[Using sum of G.P]

ny ˙ Í∵ Ì1 + ny + 2! ÍÎ ÓÔ ˙˚ ˛Ô

1 ny

0< x lim

n Æ•

n(n -1 - 1) y 2 + 2!

n




1 2

+

1 3

n Æ•

+…+

1 n

1 n

> 1+

=0 1 n

+

1 n

+…+

1 n

n n

Sn > n

and

lim n = •

n Æ•

Thus, the series is divergent. Hence, lim un = 0 is a necessary but not sufficient condition for convergence n→∞ of Sun. If lim un π 0 or lim un does not exist, then Sun is divergent. n Æ•

1.5

n Æ•

GEOMETRIC SERIES

Consider the geometric series a + ar + ar 2 +  + ar n -1 +  S n = a + ar + ar +  + ar 2

aa(1 (1 − r n ) , 1− r a (r n − 1) a( = , r −1 =

n −1

if r < 1 if r > 1

lim r n = 0

(i) When | r | < 1,

n Æ•

lim Sn =

n Æ•

a is finite. 1- r

Hence, the series is convergent. lim r n Æ ∞ (ii) When r > 1, nÆ∞

a(r n - 1) Æ∞ nÆ∞ r -1

lim Sn = lim

nÆ∞

Hence, the series is divergent. (iii) When r = 1, Sn = a + a + a +  = na lim Sn Æ ∞

nÆ∞

Hence, the series is divergent.

... (1)

1.5

Geometric Series

1.11

(iv) When r = –1, Sn = a - a + a -  ( -1)n -1 a

= 0, if n is even = a, if n is odd Hence, the series is oscillatory. (v) When r < –1, let r = –k where k > 0 a[1 - ( - k )n ] n Æ• 1+ k

lim Sn = lim

n Æ•

a[1 - ( -1)n k n ] n Æ• 1+ k = – •, if n is even = + •, if n is odd = lim

Hence, the series is oscillatory. From all the above cases, we conclude that the geometric series (1) is (i) convergent if | r | < 1 (ii) divergent if r ≥ 1 (iii) oscillatory if r £ –1 •

Note: The p series

1

 np

=

n =1

1 1

p

+

1 2

p

+

1 3p

+  is

(i) convergent if p > 1 (ii) divergent if p £ 1

Example 1 Prove that 1 +

2 4 8 16 + + + +  converges and find its sum. 3 9 27 81 [Winter 2014]

Solution The given series is geometric series with a = 1 and r = 1+

• • 2 4 8 Ê 2ˆ + + +  = Â ar n -1 = Â Á ˜ Ë ¯ 3 9 27 n =1 n =1 3

r =

2 m

un - l m

un < l+ Πvn

for all n > m

Neglecting the first m terms of Sun and Svn,

l - Œ<
(l - Œ) lim (v1 + v2 + v3 +  + vn )

n Æ•

n Æ•

for all n

1.18

Chapter 1

Sequences and Series

lim (u1 + u2 + u3 +  + un ) Æ •

[From Eq. (2)]

n Æ•

Hence, Sun is also divergent.

Example 1

•

Test the convergence of the series

Â

n=1 n

n

2

. +1

Solution un =

Let

=

vn =

Let

n n +1 1 2

3

1ˆ Ê n 2 Á1 + 2 ˜ Ë n ¯ 1 3

n2 lim

n Æ•

un = lim vn nƕ

1 1+

1 n2

=1

and Svn = Â

1 3 n2

is convergent as p =

[finite and non-zero] 3 > 1. 2

Hence, by comparison test, Sun is also convergent.

Example 2

•

Test the convergence of the series

22nn - 1

 n(n + 1)(n + 2) . n =1

Solution Let

un =

2n - 1 n(n + 1)(n + 2)

1ˆ Ê ÁË 2 - n ˜¯ = Ê 1ˆ Ê 2ˆ n2 Á 1 + ˜ Á 1 + ˜ Ë n¯ Ë n¯

Let

vn =

1 n2

1.7

Comparison Test

1.19

1ˆ Ê 2- ˜ Á Ë u n¯ lim n = lim n Æ• vn n Æ• Ê 1ˆ Ê 2ˆ ÁË 1 + n ˜¯ ÁË 1 + n ˜¯ = 2 [finite and non-zero]

and the series Svn = Â

1 n2

is convergent as p = 2 > 1.

Hence, by comparison test, Sun is also convergent.

Example 3 Test the convergence of the series

•

1

Â

n =1 1 + 2

2

+ 32 +  + n2

.

[Winter 2015] Solution 1

un =

Let

1 + 2 + 3 +  + n2 6 = n(n + 1)(2 n + 1) 6 = 1 1ˆ Ê ˆÊ n3 Á 1 + ˜ Á 2 + ˜ Ë n¯ Ë n¯

vn =

Let

2

2

1 n3

un 6 = lim n Æ• vn n Æ• Ê 1ˆ Ê 1ˆ ÁË 1 + n ˜¯ ÁË 2 + n ˜¯ = 6 [finite and non-zero] lim

1 is convergent as p = 3 > 1. n3 Hence, by comparison test, Sun is also convergent.

and Svn = Â

Example 4 •

Test the convergence of the series

Â

n+2

n =1 ( n + 1)

n

.

1.20

Chapter 1

Sequences and Series

Solution n+2

un =

Let

=

(n + 1) n 1+ 1

 1 n 2 1 +   n

vn =

Let

2 n

1 1

n2

Ê 2ˆ ÁË 1 + n ˜¯ un lim = lim n Æ• vn n Æ• Ê 1ˆ ÁË 1 + n ˜¯ = 1 [finite and non-zero] 1 1 is divergent as p = < 1. and Svn =  1 2 n2 Hence, by comparison test, Sun is also divergent.

Example 5 •

Is the series

Â

2n + 1

n =1 ( n + 1)

2

convergent or divergent?

Solution Let

un =

=

2n + 1 (n + 1)2 1ˆ Ê nÁ2 + ˜ Ë n¯ Ê 1ˆ n2 Á 1 + ˜ Ë n¯

2

1ˆ Ê 2+ ˜ 1 ÁË n¯ = n Ê 1ˆ2 ÁË 1 + n ˜¯

Let

vn =

1 n

[Summer 2015]

1.7

Comparison Test

1.21

1ˆ Ê ÁË 2 + n ˜¯ un lim = lim 2 n Æ• vn n Æ• Ê 1ˆ 1 + ÁË n ˜¯

=

2 =2 1

[finite and non-zero]

1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.

and Svn = Â

Example 6

1 •

(2 n2 - 1) 3

n =1

1 (3n3 + 2 n + 5) 4

Â

Test the convergence of the series Solution

.

1

((2 2 n2 - 1) 3

un =

Let

1

((3 3n3 + 2 n + 5) 4 2 n3

= 3 n4

1

1 ˆ3 Ê ÁË 2 - 2 ˜¯ n 1

2 5 ˆ4 Ê ÁË 3 + 2 + 3 ˜¯ n n 1

1 ˆ3 Ê ÁË 2 - 2 ˜¯ n

= 1 n 12

vn =

Let

1

2 5 ˆ4 Ê ÁË 3 + 2 + 3 ˜¯ n n

1 1 12 n 1

lim

n Æ•

un = lim vn nƕ

1 ˆ3 Ê ÁË 2 - 2 ˜¯ n 1

2 5 ˆ4 Ê ÁË 3 + 2 + 3 ˜¯ n n

1

=

(2 ) 3 1 (3) 4

[finite and non-zero]

[Winter 2013]

1.22

Chapter 1

and Svn = Â

1 1 12 n

Sequences and Series

is divergent as p =

1 < 1. 12

Hence, by comparison test, Sun is also divergent.

Example 7

1 2 3 4 + + + +. 1 3 3◊5 5◊7 7◊9 1◊

Test the convergence of the series Solution

n ((2 2 n - 1)((2 2 n + 1)

un =

Let

1 1ˆ Ê 1ˆ Ê nÁ2 - ˜ Á2 + ˜ Ë n¯ Ë n¯

=

1 n

vn =

Let

un 1 = lim n Æ• vn n Æ• Ê 1ˆ Ê 1ˆ ÁË 2 - n ˜¯ ÁË 2 + n ˜¯ lim

=

1 [finite and non - zero] 4

1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.

and Svn = Â

Example 8 Test the convergence of the series

1 3

p

+

1 5

p

+

Solution Let

un = =

Let

vn =

1 (2 n + 1) p 1 1ˆ Ê np Á 2 + ˜ Ë n¯ 1 np

p

1 7p

+ .

1.7

lim

nƕ n ƕ

un 1 = lim p n Æ• vn 1ˆ Ê ÁË 2 + n ˜¯ 1

= and Svn = Â

Comparison Test

[finite and non-zero]

2p

1

is convergent if p > 1 and divergent if p £ 1. np Hence, by comparison test, Sun is also convergent if p > 1 and divergent if p £ 1.

Example 9 Test the convergence of the series 2 + 3 + 4 + 5 +  + n + 1 + . 1 8 27 64 n3 Solution Let

un =

Let

vn =

n +1

n3 1 Ê 1ˆ = 2 Á1 + ˜ n¯ n Ë

lim

n Æ•

1 n2

un Ê 1ˆ = lim 1 + ˜ vn nÆ• ÁË n¯

=1 and Svn = Â

[finite and non-zero]

1

is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

Example 10 Test the convergence of the series Solution Let

un =

1 1+ 2

+

2 1+ 2 3

n

1+ n n +1 n = Ê ˆ 3 1 1˜ 2 Á n + 1+ Á 3 n˜ Ë n2 ¯

+

3 1+ 3 4

+.

1.23

1.24

Chapter 1

Sequences and Series

1 Ê ˆ 1 1 1˜ 2 Á n + 1+ Á 3 n˜ 2 Ën ¯

=

1

vn =

Let

1

n2

lim

n Æ•

and Svn =

1 1 n2

un 1 = lim vn nÆ• Ê ˆ Á 1 + 1+ 1 ˜ Á 3 n˜ Ë n2 ¯ = 1 [finite and non-zero]

is divergent as p =

1 < 1. 2

Hence, by comparison test, Sun is also divergent.

Example 11

1

Test the convergence of the series

2

3

+

Solution Let

n

un =

(n + 1)3 1

=

n2 3

( n + 1) 2 1

n2

= 3 n2

3

Ê 1ˆ 2 ÁË 1 + n ˜¯ 1

=

3

Ê 1ˆ 2 n Á1 + ˜ Ë n¯

Let

vn =

1 n

2 3

3

+

3 43

+.

1.7

lim

n Æ•

un = lim vn nƕ

=1 1 and Svn = Â is divergent as p = 1. n

Comparison Test

1.25

1 3

Ê 1ˆ 2 ÁË 1 + n ˜¯ [finite and non-zero]

Hence, by comparison test, Sun is also divergent.

Example 12 Test the convergence of the series

14 3

1

+

24 2

3

+

34 33

+.

Solution nth term of the numerator = a + (n – 1)d = 14 + (n – 1)10 = 10n + 4 nth term of the denominator = n3 10 n + 4 un = n3 1 Ê 4ˆ = 2 Á 10 + ˜ n¯ n Ë

vn =

Let lim

n Æ•

and Svn =

1 n2

un 4ˆ Ê = lim 10 + ˜ vn nÆ• ÁË n¯ = 10 [finite and non-zero]

1

is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

Example 13 Test the convergence of the series

1 a ◊1 + b 2

Solution Let

un =

n

a ◊ n2 + b 1 = bˆ Ê nÁa + 2 ˜ Ë n ¯

+

2 a◊2 + b 2

+

3 a ◊ 32 + b

+ .

1.26

Chapter 1

Sequences and Series

vn =

Let

1 n

un 1 = lim n Æ• vn n Æ• Ê bˆ ÁË a + 2 ˜¯ n 1 [finite and non-zero] = a lim

1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.

and Svn = Â

Example 14 Test the convergence of the series Solution Let

un =

Let

vn =

1 12 + m

+

2 22 + m

+

3 32 + m

+ .

n

n2 + m 1 = mˆ Ê n Á1 + 2 ˜ Ë n ¯ 1 n

un 1 = lim n Æ• vn n Æ• Ê mˆ ÁË 1 + 2 ˜¯ n = 1 [finite and non-zero] lim

1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.

and Svn = Â

Example 15 Test the convergence of the series Solution Let

un =

2 ◊ 13 + 5 4 ◊ 15 + 1

2 n3 + 5 4n5 + 1

+

2 ◊ 23 + 5 4 ◊ 25 + 1

+ +

2 ◊ n3 + 5 4 ◊ n5 + 1

+ .

1.7

Comparison Test

1.27

5ˆ Ê ÁË 2 + 3 ˜¯ n = 1ˆ 2Ê n Á4+ 5 ˜ Ë n ¯

vn =

Let

1 n2

Ê ÁË 2 + un lim = lim n Æ• vn n Æ• Ê ÁË 4 +

= and Svn = Â

5ˆ ˜ n3 ¯ 1ˆ ˜ n5 ¯

2 [f [finite and non-zero] 4

1

is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

Example 16 Test the convergence of the series

1 2 1◊ 3 ◊4 2

2

+

3 4 3◊ 5 ◊6 2

2

+

5 6 5◊

+. 72 ◊ 82 [Winter 2013]

Solution un =

Let

=

vn =

Let

lim

n Æ•

((2 2 n - 1) ◊ 2 n

[Using A.P.]

((2 2 n + 1)2 ((2 2 n + 2 )2 1ˆ Ê ÁË 2 - n ˜¯ ◊ 2 2

1ˆ Ê 2ˆ Ê n2 Á 2 + ˜ Á 2 + ˜ Ë n¯ Ë n¯

2

1 n2 1ˆ Ê 2Á2 - ˜ Ë n¯

un = lim 2 2 vn nÆ• Ê 1ˆ Ê 2ˆ 2 + 2 + ÁË n ˜¯ ÁË n ˜¯

=

1 [f [finite and non-zero] 4

1.28

Chapter 1

and Svn = Â

Sequences and Series

1

is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

Example 17 Test the convergence of the series

1 3 5 7 + + + +. 1◊ 2 ◊ 3 2 ◊ 3 ◊ 4 3 ◊ 4 ◊ 5 4 ◊ 5 ◊ 6 [Winter 2014]

Solution un =

Let

((22 n + 1) n(n + 1)(n + 2)

[Using A.P.]

1ˆ Ê ÁË 2 + n ˜¯ = Ê 1ˆ Ê 2ˆ n2 Á 1 + ˜ Á 1 + ˜ Ë n¯ Ë n¯ 1 vn = 2 n

Let

1ˆ Ê ÁË 2 + n ˜¯ un lim = lim n Æ• vn n Æ• Ê 1ˆ Ê 2ˆ ÁË 1 + n ˜¯ ÁË 1 + n ˜¯

= 2 [finite and non-zero] and Svn = Â

1 n2

is convergent as p = 2 > 1.

Hence, by comparison test, Sun is also convergent.

Example 18 •

Test the convergence of the series

Â(

n =1

)

n4 + 1 - n4 - 1 . [Summer 2016]

Solution Let

un = n 4 + 1 - n 4 - 1 =

( (

) ( - 1)

n4 + 1 - n4 - 1 n +1 + n 4

4

n4 + 1 + n4 - 1

)

1.7

=

Comparison Test

1.29

(n 4 + 1) - (n 4 - 1) n4 + 1 + n4 - 1 2

=

n4 + 1 + n4 - 1 1 2 = 2◊ n Ê 1 1 ˆ Á 1+ 4 + 1- 4 ˜ n n ¯ Ë vn =

Let lim

n Æ•

1 n2

un 2 = lim vn nÆ• Ê 1 1 ˆ Á 1+ 4 + 1- 4 ˜ n n ¯ Ë =2

and

1

 vn =  n2

[finite and non-zero]

is convergent as p = 2 > 1.

Hence, by comparison test, Sun is also convergent.

Example 19 Check for convergence of the series

•

5n3 - 3n

n =1

n2 (n - 2)(n2 + 5)

Â

.

[Winter 2016] Solution un =

5n 3 - 3n n2 (n - 2)(n2 + 5)

3ˆ Ê n3 Á 5 - 2 ˜ Ë n ¯ = 2ˆ Ê 5ˆ Ê n5 Á 1 - ˜ Á 1 + 2 ˜ Ë n¯ Ë n ¯ 3ˆ Ê ÁË 5 - 2 ˜¯ n = 5ˆ 2ˆ Ê 2Ê n Á1 - ˜ Á1 + 2 ˜ Ë n¯ Ë n ¯

1.30

Chapter 1

Sequences and Series

vn =

Let

1 n2

3ˆ Ê 5- 2˜ Á Ë u n ¯ lim n = lim n Æ• vn n Æ• Ê 2ˆ Ê 5ˆ ÁË 1 - n ˜¯ ÁË 1 + 2 ˜¯ n =5 and

1

 vn =  n2

[finite and non zero]

is convergent as p = 2 > 1.

Hence, by comparison text,

 un

is also convergent.

Example 20 •

Test the convergence of the series Solution Let

È1

Ê n + 1ˆ ˘ ˜ . n ¯ ˙˚

Â Í n - log ÁË

n =1 Î

un =

1 Ê n + 1ˆ - log Á Ë n ˜¯ n

=

1 Ê 1ˆ - log Á 1 + ˜ Ë n n¯

=

1 Ê1 1 1 1 ˆ - Á - 2 + 3 - 4 + ˜ ¯ n Ë n 2n 3n 4n

1 1 1 + - 2 n 2 3n 3 4 n 4 1 Ê1 1 1 ˆ = 2Á + 2 - ˜ Ë ¯ 2 3 n n 4n =

vn =

Let lim

n Æ•

1 n2

un 1 Ê1 1 ˆ = lim + - ˜ ¯ vn nÆ• ÁË 2 3n 4 n2

=

1 [f [finite and non-zero] 2

1 is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

and Svn = Â

1.7

Example 21

•

Test the convergence of the series

Comparison Test

1.31

1

 sin n .

n =1

Solution un = sin

Let

=

1 n

1 1 1 + - 3 n 3! n 5! n5

1Ê 1 1 ˆ 1+ - ˜ Á 2 4 ¯ n Ë 3! n 5! n 1 vn = n =

Let lim

n Æ•

un 1 1 Ê ˆ = lim Á 1 + - ˜ 2 4 n Æ• Ë ¯ vn 3! n 5! n =1

[finite and non-zero]

1 is divergent as p = 1. n Hence, by comparison test, Sun is also divergent.

and Svn = Â

Example 22 È Test the convergence of the series Â Í n3 + 1 n =1 Í Î Solution •

Let

(

)

1 3

˘ - n ˙. ˙˚

[Summer 2017]

1 È ˘ un = Í(n3 + 1) 3 - n ˙ ÍÎ ˙˚

1

1 ˆ3 Ê = n Á1 + 3 ˜ - n Ë n ¯ È 1Ê1 ˆ 1Ê1 ˆÊ1 ˆ Í 1 1 3 ÁË 3 - 1˜¯ 1 2 3 ÁË 3 - 1˜¯ ÁË 3 - 2˜¯ Ê ˆ = n Í1 + ◊ 3 + ÁË 3 ˜¯ + ÍÎ 3 n 2! 3! n

=

1 1 1 1 5 1 ◊ 2 - 2 ◊ 5 + 4 ◊ 8 - 3 n 3 n 3 n

˘ ˙ Ê 1ˆ ÁË 3 ˜¯ + ˙˙ - n n ˚ 3

1.32

Chapter 1

Sequences and Series

1 n

lim

n Æ•

-

1

1

2

3

n

+

5

◊

4

1

-

n6

1

vn

Let

1

2

n2

un 1 1 = n Æ• vn 3 32

1 3

1 3

5

+

4

◊

1

-

6

and non-zero]

and the series Svn = Â

1 n2

is convergent as p = 2 > 1.

Hence, by comparison test, Sun is also convergent.

Example 23

Â

Test the convergence of the series

È

1

1

.

n 1

Solution 1

Let

1 3

un

1 3

1

=n

˘ ˙ ˙

1

=n

=

=

Let

vn

3

1 2 3n 3

-

1 5 9n 3

2!

+

5 8 81n 3

-

1 Ê1 1 5 ˆ + - ˜ 2 Á Ë 3 9n 81n2 ¯ n3 1 2

n3

˘ ˙ ˙ ˙

1.7

Comparison Test

1.33

un 5 Ê1 1 ˆ = lim Á + - ˜ 2 n Æ• vn n Æ• Ë 3 ¯ 9n 81n lim

= and Svn = Â

1 2 n3

1 3

[finite and non-zero] 2 < 1. 3

is divergent as p =

Hence, by comparison test, Sun is also divergent.

Example 24 ∞

Test the convergence of the series

∑ n =1

n2 + n + 1 − n2 − n + 1 . n

Solution Let

un =

n2 + n + 1 - n2 - n + 1 n 1

1

È Ê 1 1 ˆ ˘2 È Ê 1 1 ˆ ˘2 = Í1 + Á + 2 ˜ ˙ - Í1 + Á - + 2 ˜ ˙ Î Ë n n ¯˚ Î Ë n n ¯˚

Expanding using binomial expansion, È ˘ 1Ê1 ˆ ÁË 2 - 1˜¯ 1 1 2 Í 1 1 1 ˙ 2 Ê ˆ Ê ˆ ˙ un = Í1 + Á + 2 ˜ + + +  ÁË n ˜ Í 2Ën n ¯ 2! ˙ n2 ¯ Í ˙ Î ˚ È ˘ 1Ê1 ˆ 2 ÁË 2 - 1˜¯ Í 1 ˙ 1 1 1 1 2 Ê ˆ Ê ˆ - Í1 + Á - + 2 ˜ + - + 2 ˜ + ˙ Á Ë n n ¯ Í 2Ë n n ¯ 2! ˙ Í ˙ Î ˚

È ˘ 1 1 1Ê 1 2 1ˆ = Í1 + + 2 - Á 2 + 3 + 4 ˜ + ˙ Ë ¯ 2 n 8 2n n n n Î ˚ È ˘ 1 1 1Ê 1 2 1ˆ - Í1 + 2 - Á 2 - 3 + 4 ˜ + ˙ Ë ¯ 2 n 8 2n n n n Î ˚

1.34

Chapter 1

Sequences and Series

=

1 1 + n 2 n3

=

1Ê 1 ˆ 1 - 2 + ˜ Á Ë ¯ n 2n

vn =

Let

1 n

un 1 Ê ˆ = lim Á 1 - 2 + ˜ n Æ• vn n Æ• Ë ¯ 2n lim

=1 and Svn = Â

[finite and non-zero]

1 is divergent as p = 1. n

Hence, by comparison test, Sun is also divergent.

Example 25 Ê n2 + 1 - n ˆ Test the convergence of the series  Á ˜. ÁË ˜¯ np Solution Let

un =

n2 + 1 - n np

1 È ˘ 1 ˆ2 ˙ Ê Í n Á1 + 2 ˜ - 1 ÍË ˙ n ¯ ÍÎ ˙˚ = p n ÈÏ ¸ ˘ 1Ê1 ˆ 1Ê1 ˆÊ1 ˆ -1 -1 - 2˜ Í Ô ˙ ¯ 1 n Í ÔÔ 1 2 ÁË 2 ˜¯ 1 2 ÁË 2 ˜¯ ÁË 2 Ô = p Ì1 + 2 + ◊ 4 + ◊ 6 + ˝ - 1˙ 2! 3! n Í Ô 2n n n Ô ˙˙ ÍÔ Ô˛ ˚ ÎÓ

=

n Ê 1 1 1 ˆ + - ˜ Á ¯ n p Ë 2 n2 8n 4 16 n6

=

1 Ê1 1 1 ˆ - ˜ ÁË 2 - 2 + 4 ¯ n 8n 16 n p +1

1.7

Let

Comparison Test

1.35

1

vn =

p +1

n un 1 1 Ê1 ˆ lim = lim Á - 2 + - ˜ 4 n Æ• vn n Æ• Ë 2 ¯ 8n 16 n 1 = [finite and non-zeero] 2

and Svn = Â

1 n

p +1

is convergent if p + 1 > 1, i.e., p > 0 and divergent if p + 1 £ 1,

i.e., p £ 0. Hence, by comparison test, Sun is also convergent if p > 0 and divergent if p £ 0.

Example 26 Test the convergence of the series where x is a positive fraction.

1 1 1 1 1 + + + + +, x x -1 x +1 x - 2 x + 2

Solution Since it is an infinite series, by ignoring the first term, the series can be rewritten as Ê 1

1 ˆ

Ê 1

1 ˆ

 un = ÁË x - 1 + x + 1˜¯ + ÁË x - 2 + x + 2 ˜¯ +  =

2x 2x + + x 2 - 12 x 2 - 22 2x

=Â

x 2 - n2 2x

un =

x 2 - n2 2x = Ê x2 ˆ n2 Á 2 - 1˜ Ën ¯

vn =

Let lim

n Æ•

1 n2

un 2x = lim vn nÆ• Ê x 2 ˆ Á 2 - 1˜ Ën ¯ = –2x

[finite and non-zero]

1.36

Chapter 1

and Svn =

Sequences and Series

1

is convergent as p = 2 > 1. n2 Hence, by comparison test, Sun is also convergent.

EXERCISE 1.2 1. Test the convergence of the following series: (i) (iii) (v) (vii)

∑n ∑(

2

1 +1

)

n4 + 1 − n4 − 1

n

∑(

(iv)

∑ 

(vi)

∑

(viii)

∑

p

∑ (n + 1)

q

∑ tan

(ii)

−1

 1   n

n +1− n



)

np n +1+ n

 

 1 tan    n n

1

1 n

 b a+   a+ n

.

 Ans. :    (i) Convergent (ii) Divergent ( iii ) Convergent     1 1 ntt if p < − D Divergent ivergent if p ≥ − (iv) Convergen  2 2   Dive ergen rgent if p − q + 1 ≥ 0   (v) Convergent if p − q + 1 < 0, Div (vi) Convergent  (vii) Divergent   (viii) Convergent if a > 1, D  Divergent ivergent if a ≤ 1 2. Test the convergence of the series

((1 1+ 1 + a)( )(1 + b) ((2 2 + a)( )(2 + b) ((3 3 + a)( )(3 + b) + + +…. 1⋅ 2 ⋅ 3 2⋅3⋅4 3⋅ 4 ⋅5 [Ans. : Divergent]

1.8

D’ALEMBERT’S RATIO TEST

If Sun is a positive-term series and lim

n Æ•

(i) Sun is convergent if l < 1 (ii) Sun is divergent if l > 1

un +1 = l then un

1.8

D’Alembert’s Ratio Test

1.37

Proof

un +1 = l < 1. n Æ• un

Case I If lim

Consider a number l < r < 1 such that

un +1 < r for all n > m un

… (1)

Neglecting the first m terms, •

Â

un = um ++11 + um + 2 + um + 3 + … •

n m +1 n=

Ê u ˆ u u = um +1 Á 1 + m + 2 + m + 3 + m + 4 + …˜ um +1 um +1 um +1 Ë ¯

Ê u ˆ u u u u u = um +1 Á 1 + m + 2 + m + 3 ◊ m + 2 + m + 4 ◊ m + 3 ◊ m + 2 + …˜ um +1 um + 2 um +1 um + 3 um + 2 um +1 Ë ¯ < um +1 +1 (1 + r + r ◊ r + r ◊ r ◊ r + …)

[[Usin Using Eq. (1)]

= um +1 +1 (1 + r + r + r + …) 2

= um + 1 ◊ •

\

Â

un
1, n Æ• un lim

un +1 > 1 for all n > m un

... (2)

1.38

Chapter 1

Sequences and Series

Neglecting the first m terms, •

Â

un

um

m

m+4

m

n m +1

Ê

= um

Ê

= um

+

•

+

ˆ um + 2 u u + + +4 + … um +1 um +1 um +1

+

um + 2 u u u u u + ◊ m+2 + m+ 4 m+3 +… um +1 um + 2 um +1 um + 3 um + 2 um +1

> um +1 (1+ + + + …) \

o

Sn

m

lim

n m

1

n Æ•

n

terms)

is positive]

•

Thus, the series

Â

un is divergent.

n m +1

The nature of a series remains unchanged if we neglect a finite number of terms in the •

beginning. Hence, the series

 un

is divergent.

n

un +1 = 1 the ratio test fails, i.e. no conclusion can be drawn about the n Æ• un

Note 1: If lim

convergence or divergence of the series.

Note 2: It is convenient to use D’Alembert’s ratio test in the following form: u If Sun is a positive term series and lim n = l, then n Æ• un +1 (i) Sun is convergent if l > 1 (ii) Sun is divergent if l < 1 (iii) The ratio test fails if l = 1

Example 1 Test the convergence of the series

•

32 n

n£0

23 n

Â

Solution Let

un =

32 n 23 n

.

[Summer 2014]

1.8

32( n +1)

un +1 =

23( n +1)

=

D’Alembert’s Ratio Test

1.39

32 n + 2 23 n + 3

un 32 n 23n + 3 = 3n ◊ 2 n + 2 un +1 2 3 =

lim

n Æ•

23 2

3

=

8 9

un 8 8 = lim = < 1 un +1 nƕ 9 9

Hence, by D’Alembert’s ratio test, the series is divergent.

Example 2 5n -1 . Â n =1 n ! •

Test the convergence of the series Solution

5n -1 n! 5n = (n + 1)!

un =

Let

un +1

n -1 1 un 5n(n + 1)! = ◊ un +1 n! 5n

n +1 5 n +1 = lim Æ • >1 n Æ• 5 =

lim

n Æ•

un un +1

Hence, by D’Alembert’s ratio test, the series is convergent.

Example 3 •

Test the convergence of the series

Â

2n

. 3 n =1 n + 1

Solution Let

un = un +1 =

2n n3 + 1 2 n +1 (n + 1)3 + 1

[Winter 2016]

1.40

Chapter 1

Sequences and Series

un 2 n (n + 1)3 + 1 = 3 ◊ un +1 n + 1 2 n +1 3

1 Ê 1ˆ ÁË 1 + n ˜¯ + 3 1 n = ◊ 1 2 1+ 3 n 3

1 Ê 1ˆ 1+ ˜ + 3 Á Ë u n¯ n 1 lim n = lim ◊ 1 n Æ• un +1 n Æ• 2 1+ 3 n 1 = 1 Hence, by D’Alembert’s ratio test, the series is convergent.

1.8

D’Alembert’s Ratio Test

Example 5 Test the convergence of the series

n !(2) n ∑ nn .

Solution un =

Let

un +1 =

n !(2)n nn (n + 1)!( )!(2)n +1 (n + 1)n +1

un n ! 2 n (n + 1)n +1 = n ◊ un +1 n (n + 1)! )! 2 n +1 =

1 Ê n + 1ˆ 2 ÁË n ˜¯

n

un 1 Ê 1ˆ = lim Á 1 + ˜ nnÆ• Æ• un +1 n Æ• Ë n¯ 2

n

lim

e >1 2 Hence, by D’Alembert’s ratio test, the series is convergent. =

Example 6 •

Test the convergence of the series

n3 . Â n =1 ( n - 1)!

Solution Let

un =

n3 (n -1 - 1)!

un +1 =

(n + 1)3 n!

un n3 n! = ◊ un +1 (n -1 - 1)! (n + 1)3 = =

n3 n(n -1 - 1)! ◊ (n -1 - 1)! (n + 1)3 n Ê 1ˆ ÁË 1 + n ˜¯

3

1.41

1.42

Chapter 1

Sequences and Series

lim

n Æ•

un n = lim Æ • >1 un +1 nÆ• Ê 1 ˆ 3 ÁË 1 + n ˜¯

Hence, by D’Alembert’s ratio test, the series is convergent.

Example 7

∞

(n + 1) n Test the convergence of the series ∑ . n! n =1 Solution (n + 1)n n! (n + 2)n +1 = (n + 1)!

un =

Let

un +1

un (n + 1)n (n + 1)! = ◊ un +1 n! (n + 2)n +1 = =

lim

n Æ•

(n + 1)n n!

((n n + 1)(n !)

[(n + 1) + 1]n+1

1 1 ˆ Ê ÁË 1 + n + 1˜¯

n +1

un 1 = lim n +1 un +1 nÆ• Ê 1 ˆ 1 + ÁË n + 1˜¯

1 1 2 Hence, by D’Alembert’s ratio test, the series is convergent.

Example 9 •

Test the convergence of the series Solution

Â

n =1 3

un =

Let

5n + a

un +1 =

n

+b

, a > 0, b > 0. 0

5n + a 3n + b 5n +1 + a 3n +1 + b

un 5n + a 3n +1 + b = lim n ◊ n Æ• un +1 n Æ• 3 + b 5n +1 + a lim

a b 3+ n n 5 3 = lim ◊ b a n Æ• 1+ n 5 + n 3 5 3 = 1 1 n Æ• un +1 n Æ• 1+ n Hence, by D’Alembert’s ratio test, the series is convergent. lim

Example 13 2 p 3p 4 p Test the convergence of 1 + + + + … ,( p > 0). 2! 3! 4!

Solution un =

Let

un +1 =

lim

nnƕ ƕ

un un +1

np n! (n + 1) p (n + 1)!

np n! = lim n Æ• ( n + 1) p (n + 1)! = lim

n Æ•

= lim

n Æ•

np (n + 1)

p

◊

(n + 1) Ê 1ˆ ÁË 1 + n ˜¯

p

(n + 1)! n!

Æ • >1

Hence, by D’Alembert’s ratio test, the series is convergent.

Example 14 Test the convergence of the series

1 2 3 4 + + + + . 2 3 1 + 2 1 + 2 1 + 2 1 + 24

1.46

Chapter 1

Sequences and Series

Solution un =

Let

un +1 = un un +1

n 1 + 2n n +1

1 + 2 n +1 n 1 + 2 n +1 = ◊ 1 + 2n n + 1 1 +2 2n = Ê 1 ˆ Ê 1ˆ ÁË n + 1˜¯ ÁË 1 + n ˜¯ 2

1 +2 un 2n lim = lim n Æ• un +1 n Æ• Ê 1 ˆ Ê 1ˆ ÁË n + 1˜¯ ÁË 1 + n ˜¯ 2

=2>1

È∵ lim 2 n Æ • ˘ ÍÎ nÆ• ˙˚

Hence, by D’Alembert’s ratio test, the series is convergent.

Example 15 •

Test the convergence of the series

2 ◊ 4 ◊ 6 … 2n

 5 ◊ 8 ◊11…(3n + 2).

n =1

Solution Let

2 ◊ 4 ◊ 6… 6 … 2n 5 ◊ 8 ◊ 11… (3n + 2) 2 ◊ 4 ◊ 6… 6 … 2 n(2 n + 2) = 5 ◊ 8 ◊ 11… (3n + 2)(3n + 5)

un = un +1

un 2 ◊ 4 ◊ 6… 6 … 2n 5 ◊ 8 ◊ 11… (3n + 2)(3n + 5) = ◊ un +1 5 ◊ 8 ◊ 11… (3n + 2) 2 ◊ 4 ◊ 6…2 2n n(2 n + 2) 33n + 5 2n + 2 2n 5 3+ n = 2 2+ n =

[Using A.P.]

1.8

D’Alembert’s Ratio Test

1.47

5 3+ un n lim = lim 2 n Æ• un +1 n Æ• 2+ n 3 = >1 2 Hence, by D’Alembert’s ratio test, the series is convergent.

Example 16 Test the convergence of the series Solution Let

un =

2 2 ◊ 5 2 ◊ 5 ◊ 8 2 ◊ 5 ◊ 8 ◊ 111 + + + + . 1 1 ◊ 5 1 ◊ 5 ◊ 9 1 ◊ 5 ◊ 9 ◊ 113

2 ◊ 5 ◊ 8 ◊ 11… (3n -1 - 1) 1 ◊ 5 ◊ 9 ◊ 13… (4 n - 3)

[Using A.P.]

2 ◊ 5 ◊ 8 ◊ 11… (3n -1 - 1)(3n + 2) 1 ◊ 5 ◊ 9 ◊ 13… (4 n - 3)(4 n + 1) 2 ◊ 5 ◊ 8 ◊ 11… 1… (3n - 1) un 1 ◊ 5 ◊ 9 ◊ 13… 3… (4 n - 3) lim = lim n Æ• un +1 n Æ• 2 ◊ 5 ◊ 8 ◊ 11…(((3n - 1))(((3n + 2) 1 ◊ 5 ◊ 9 ◊ 13 13… 3… (4 n - 3)( )(4 n + 1) 44nn + 1 = lim n Æ• 3 3n + 2 1 4+ n = lim 2 n Æ• 3+ n 4 = >1 3 Hence, by D’Alembert’s ratio test, the series is convergent. un+1 =

Example 17

2

2

2

Ê 1ˆ Ê 1◊ 2 ˆ Ê 1◊ 2 ◊ 3 ˆ +Á + … •. • Test the convergence of the series Á ˜ + Á ˜ Ë 3¯ Ë 3 ◊ 5¯ Ë 3 ◊ 5 ◊ 7 ˜¯ Solution Let

È 1 ◊ 2 ◊ 3… n ˘ un = Í ˙ Î 3 ◊ 5 ◊ 7 … (2 n + 1) ˚

2

È ˘ 1 ◊ 2 ◊ 3 … n(n + 1) un +1 = Í ˙ Î 3 ◊ 5 ◊ 7 … (2 n + 1)(2 n + 3) ˚

[Using A.P.] 2

1.48

Chapter 1

Sequences and Series

È 1 ◊ 2 ◊ 3… n ˘ Í 3 ◊ 5 ◊ 7 … ((2 n + 1) ˙ Î ˚

2

un = lim 2 n Æ• un +1 n Æ• È ˘ 1 ◊ 2 ◊ 3… 3… n n((n + 1) Í 3 5 ◊ 7 … (2 n + 1)((2 2 n + 3) ˙˚ Î lim

2 + 3˘ È 2n = lim Í n Æ• Î n + 1 ˙ ˚ 3˘ È Í2 + n ˙ = lim Í ˙ n Æ• Í1+ 1 ˙ ÍÎ n ˙˚ = 4 >1

2

2

Hence, by D’Alembert’s ratio test, the series is convergent.

Example 18 Test the convergence of the series 2 +

3 4 5 x + x 2 + x 3 + . 2 3 4 [Summer 2014]

Solution Let

n + 1 n -1 x n n+2 n = x n +1

un = un +1

un (n + 1) x n -1 n +1 = ◊ un +1 n ( n + 2) x n =

1 1 1+ n n ◊1 2 1 x 1+ n

1+

1 1 1+ 1+ un n◊ n ◊1 lim = lim 2 x n Æ• un +1 n Æ• 1 1+ n 1 = x

1.8

D’Alembert’s Ratio Test

By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x 1 The test fails if = 1, i.e., x = 1. x n +1 1 = 1+ n n Ê 1ˆ lim un = lim Á 1 + ˜ n Æ• n Æ• Ë n¯ un =

Then

=1

[finite and non-zero]

Sun is divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.

Example 19 •

Test the convergence of the series

Â

(n + 1)

n =1

Solution un =

Let

un +1 =

n +1 n2

n

2

xn.

xn

n+2 (n + 1)2

x n +1

un (n + 1) x n (n + 1)2 = ◊ un +1 n2 (n + 2) x n +1 =

(n + 1)3 n2 ((nn + 2) x 3

Ê 1ˆ ÁË 1 + n ˜¯ 1 = ◊ Ê 2ˆ x 1 + ÁË n ˜¯ 3

lim

n Æ•

un un +1

Ê 1ˆ ÁË 1 + n ˜¯ 1 = lim ◊ n Æ• Ê 2ˆ x ÁË 1 + n ˜¯ =

1 x

1.49

1.50

Chapter 1

Sequences and Series

By D’Alembert’s ratio test, the series is (i) convergent if

1 > 1 or x < 1 x

(ii) divergent if

1 < 1 or x > 1 x

The test fails if

1 = 1, i.e., x = 1. x un =

Then

vn =

n2

1 Ê 1ˆ 1+ ˜ n ÁË n¯

= Let

n +1

1 n

un  1 = lim lim 1 +  n →∞ v n →∞  n n lim m

=1

[finite and non-zero]

1 is divergent as p = 1. n By comparison test, Sun is also divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.

and Svn = Â

Example 20 Ê xn ˆ Test the convergence of the series  Á 2 ˜ , for x > 0. n =1 Ë n + 1¯ Solution •

Let

un =

un +1 =

xn n2 + 1

x n +1 (n + 1)2 + 1

un x n (n + 1)2 + 1 = 2 ◊ un +1 n + 1 x n +1 2

1 Ê 1ˆ ÁË 1 + n ˜¯ + 2 1 n = ◊ 1 x 1+ 2 n

1.8

D’Alembert’s Ratio Test

2

un n Æ• un +1 lim

1 Ê 1ˆ ÁË 1 + n ˜¯ + 2 1 n = lim ◊ 1 n Æ• x 1+ 2 n 1 = x

By D’Alembert’s ratio test, the series is (i) convergent if

1 > 1 or x < 1 x

1 < 1 or x > 1 x The test fails if x = 1.

(ii) divergent if

1

un =

Then

n +1 1 = 1ˆ Ê n2 Á 1 + 2 ˜ Ë n ¯ 1 vn = 2 n un 1 lim = lim n Æ• vn n Æ• Ê 1ˆ ÁË 1 + 2 ˜¯ n

Let

2

=1 and Svn = Â

1 n2

[finite and non-zero]

is convergent as p = 2 > 1.

By comparison test, Sun is also convergent if x = 1. Hence, the series is convergent for x £ 1 and is divergent for x > 1.

Example 21 Test the convergence of the series

Â

Solution Let

un =

2n n +1 4

2n n4 + 1

x n , x > 0.

xn

1.51

1.52

Chapter 1

Sequences and Series

2 n +1

un +1 = un un +1

(n + 1)4 + 1

x n +1

2 n x n (n + 1)4 + 1 ◊ n 4 + 1 2 n +1 x n +1

=

4

1 Ê 1ˆ ÁË 1 + n ˜¯ + 4 n = 1ˆ Ê ÁË 1 + 4 ˜¯ 2 x n 4

un n Æ• un +1 lim

1 Ê 1ˆ ÁË 1 + n ˜¯ + 4 n = lim n Æ• Ê 1ˆ ÁË 1 + 4 ˜¯ 2 x n 1 = 2x

By D’Alembert’s ratio test, the series is (i) convergent if 1 > 1 or x < 1 2x 2 (ii) divergent if 1 < 1 or x > 1 2x 2 The test fails if 1 = 1 or x = 1 . 2x 2 Then

un = =

2n

◊

1

n + 1 2n 1 4

n4 + 1 1 = 1ˆ Ê n4 Á1 + 4 ˜ Ë n ¯

vn =

Let lim

n Æ•

1 n4

un 1 = lim 1ˆ vn nÆ• Ê ÁË 1 + 4 ˜¯ n

=1 and Σvn = ∑

1 is convergent as p = 4 > 1. n4

[finite and non-zero]

1.8

1 . 2

By comparison test, Sun is also convergent if x = Hence, the series is convergent for x £

D’Alembert’s Ratio Test

1 1 and is divergent for x > . 2 2

Example 22

∑

Test the convergence of the series

n n +1 2

xn .

Solution un =

Let

un +1 = un = un +1 =

n n +1 2

◊ xn

(n + 1) (n + 1)2 + 1 n n +1 2

◊ xn

◊ x n +1 (n + 1)2 + 1 1 ◊ n +1 n +1 x

n (n2 + 2 n + 2) 1 ◊ ◊ (n + 1) x (n2 + 1)

Ê 2 2ˆ ÁË 1 + n + 2 ˜¯ un 1 n lim = lim ◊ nÆ• n Æ• un +1 n Æ• Ê 1ˆ Ê 1ˆ x ÁË 1 + n ˜¯ ÁË 1 + 2 ˜¯ n

1 x By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1, or x < 1 x =

(ii) divergent if

1 < 1, orr x >1 x

The test fails for x = 1. Then

un =

n n2 + 1 1

=

n2 n 1+

1 n2

1.53

1.54

Chapter 1

Sequences and Series

1

=

1 n2

1 n2

1

vn =

Let

1+

1

n2 un = lim n Æ• vn n Æ• lim

and Svn = Â

1 1 n2

1

1 n2 = 1 [[finite finite and non-zero]

is divergent for p =

1+

1 1 or x < 1 x

1 < 1 or x > 1 x

1 = 1 or x = 1. x 1 1+ n 1 = 1  n  + 1 n 

Then

un =

Let

vn =

1 n

un 1 = lim n Æ• vn n Æ• Ê 1 ˆ ÁË n + 1˜¯ = 1 [finite and non-zero] lim

and Σvn = ∑

1 is divergent as p = 1. n

By comparison test, Sun is also divergent for x = 1. Hence, the series is convergent for x < 1 and is divergent for x ≥ 1.

Example 24 Test the convergence of the series Solution Let

x x2 x3 x4 + + + +. 1◊ 2 3 ◊ 4 5 ◊ 6 7 ◊ 8

un =

un +1 =

xn ((22 n - 1)2 n

x n +1 (2 n + 1)(2 n + 2)

un xn (2 n + 1)(2 n + 2) = ◊ un +1 (2 n - 1)2 n x n +1 1ˆ Ê ÁË 2 + n ˜¯ = 1ˆ Ê ÁË 2 - n ˜¯

Ê 1ˆ 1 ÁË 1 + n ˜¯ ◊ x

1.55

1.56

Chapter 1

Sequences and Series

1ˆ Ê ÁË 2 + n ˜¯ un lim = lim n Æ• un +1 n Æ• Ê 1ˆ ÁË 2 - n ˜¯ =

Ê 1ˆ 1 ÁË 1 + n ˜¯ ◊ x

1 x

By D’Alembert’s ratio test, the series is (i) convergent if (ii) divergent if The test fails if

1 > 1 or x < 1 x

1 < 1 or x > 1 x

1 = 1 or x = 1. x

1 (2n − 1)2n

un =

Then

1

=

vn =

Let lim m

n →∞

1  2 2 2−  2n  n

1 n2

un 1 = lim 1 vn n→∞  22 −   n =

1 [[finite finite and non-zero] 4

1 is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent.

and Σvn = ∑

Hence, the series is convergent for x £ 1 and is divergent for x > 1.

Example 25 Test the convergence of the series

x x2 1 + + +. 1◊ 2 ◊ 3 4 ◊ 5 ◊ 6 7 ◊ 8 ◊ 9 [Summer 2017]

1.8

D’Alembert’s Ratio Test

Solution un =

Let

xn - 1 (3n - 2)(3n - 1) 3n xn (3n + 1) (3n + 2)(3n + 3)

un + 1 =

un (3n + 1) (3n + 2) (3n + 3) xn - 1 = ◊ un + 1 (3n - 2) (3n - 1) (3n) xn 1ˆ Ê 2ˆ Ê 3ˆ Ê ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ 1 = ◊ 2ˆ Ê 1ˆ x Ê 3 3 3 ÁË n ˜¯ ÁË n ˜¯ 1ˆ Ê 2ˆ Ê 3ˆ Ê ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ ÁË 3 + n ˜¯ 1 un lim = lim ◊ n Æ • un + 1 nÆ• 2ˆ Ê 1ˆ x Ê 3 3 3 ÁË n ˜¯ ÁË n ˜¯ 1 x By D¢ Alembert’s ratio text, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x =

The test fails if

1 = 1 or x = 1 . x

Then

un = =

vn =

Let lim

n Æ•

1 (3n - 2)(3n - 1)(3n) 1 2ˆ Ê 1ˆ Ê n3 Á 3 - ˜ Á 3 - ˜ 3 Ë n¯ Ë n¯ 1 n3

un 1 = lim 2ˆ Ê 1ˆ vn n Æ• Ê ÁË 3 - n ˜¯ ÁË 3 - n ˜¯ 3 =

1 27

[finite and non-zero]

1.57

1.58

Chapter 1

and

 vn =  n3

Sequences and Series

1

is convergent as p = 3 > 1.

 un

By comparison test,

is also convergent for x = 1.

Hence, the series is convergent for x £ 1 and is divergent for x > 1.

Example 26 3 4 (n + 1) Test the convergence of the series 2 x + x 2 + x 3 +  + 3 x n + . 8 27 n Solution un =

Let

un +1

n +1

◊ xn n3 n+2 = ◊ x n +1 (n + 1)3

un (n + 1) x n (n + 1)3 = ◊ un +1 n3 (n + 2) x n +1 4

Ê 1ˆ ÁË 1 + n ˜¯ 1 = ◊ Ê 2ˆ x ÁË 1 + n ˜¯ 4

lim

n Æ•

un un +1

Ê 1ˆ ÁË 1 + n ˜¯ 1 = lim ◊ n Æ• Ê 2ˆ x 1 + ÁË n ˜¯ =

1 x

By D’Alembert’s ratio test, the series is (i) convergent if (ii) divergent if The test fails if Then

1 > 1 or x < 1 x

1 < 1 or x > 1 x

1 = 1 or x = 1 . x un =

n +1

n3 1 Ê 1ˆ = 2 Á1 + ˜ n¯ n Ë

1.8

vn =

Let lim

n Æ•

1.59

1 n2

un Ê 1ˆ = lim 1 + ˜ vn nÆ• ÁË n¯

=1 and Svn = Â

D’Alembert’s Ratio Test

[finite and non-zero]

1

is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent for x = 1. Hence, the series is convergent for x £ 1 and is divergent for x > 1.

Example 27 x2 x4 x6 Test the convergence of the series + + + + . 2 1 3 2 4 3 5 4 [Winter 2013] Solution 1

un =

Let

u n +1 =

n−2 x 2 n−

(n + 1) n

x2n ( n + 2) n + 1

n−2 un x 2 n− ( n + 2) n + 1 = ⋅ un +1 (n + 1) n x2n

=

( n + 2) n + 1 1 ⋅ (n +1 + 1) n x2

Ê 2ˆ ÁË 1 + n ˜¯ 1 1 = 1+ ◊ 2 n x Ê 1ˆ ÁË 1 + n ˜¯

lim

n Æ•

un un +1

Ê 2ˆ ÁË 1 + n ˜¯ 1 1 = lim 1+ ◊ 2 n Æ• Ê 1ˆ n x ÁË 1 + n ˜¯ =

By D’Alembert’s ratio test, the series is 1 (i) convergent if 2 > 1 or x2 < 1 x 1 2 (ii) divergent if 2 < 1 or x > 1 x

1 x2

1.60

Chapter 1

The test fails if

Sequences and Series

1 x

2

= 1 or x 2 = 1.

un =

Then

=

vn =

Let

1 (n + 1) n 1 3 n2

Ê 1ˆ ÁË 1 + n ˜¯

1 3

n2

un 1 = lim vn nÆ• Ê 1 ˆ ÁË 1 + n ˜¯ = 1 [finite and non-zero] 3 1 and Svn = 3 is convergent as p = > 1. 2 n2 By comparison test, Sun is also convergent for x2 = 1. lim

n Æ•

Hence, the series is convergent for x2 £ 1 and is divergent for x2 > 1.

Example 28 Test the convergence of the series 1 +

3 5 7 3 9 4 x + x2 + x + x + . 2 9 28 65

Solution un =

Let

un n Æ• un +1 lim m

xn

n +1 2n + 3 3

x n +1 (n + 1)3 + 1 (2 n + 1) x n [(n + 1 1))3 + 1] = ◊ n 3 + 1 (2 2n n + 3) x n +1

un +1 = un un +1

2n + 1

3 1 ˆ ÈÊ 1 ˆ 1˘ Ê 2 + 1 + + 3˙ Í ÁË ˜ Á ˜ n ¯ ÍË n¯ n ˙˚ Î = 3ˆ Ê 1ˆ Ê ÁË 2 + n ˜¯ ÁË 1 + 3 ˜¯ x n 3 È 1ˆ Ê 1ˆ 1˘ Ê ÁË 2 + n ˜¯ ÍÁË 1 + n ˜¯ + 3 ˙ n ˙˚ ÍÎ = lim n Æ• 3ˆ Ê 1ˆ Ê ÁË 2 + n ˜¯ ÁË 1 + 3 ˜¯ x n

[Neglecting the first term]

1.8

D’Alembert’s Ratio Test

1.61

1 x By D’Alembert’s ratio test, the series is =

(i) convergent if

1 > 1 or x < 1 x

1 < 1 or x > 1 x The test fails if x = 1. Then

(ii) divergent if

un =

2n + 1 n3 + 1

1 n = 1 2  n 1 + 3   n  2+

vn =

Let

1 n2

1 2+ un n lim m = lim n →∞ v n →∞  1  n 1 + 3  n

=2 and Svn = Â

[finite and non-zero]

1

is convergent as p = 2 > 1. n2 By comparison test, Sun is also convergent if x = 1.

Hence, the series is convergent for x £ 1 and is divergent for x > 1.

EXERCISE 1.3 Test the convergence of the following series: 1. 1 +

2 2 32 4 2 + + + …∞ 2! 3! 4!

[Ans.: Convergent]

∞

2.

4 ⋅ 7 ⋅ 10 …(3n + 1) 1⋅ 2 ⋅ 3 ⋅ 4 … n n =1

∑

1 2 3 + + + …∞ 3. 2 1 5 1+ 5 1+ 1 + 53 4. 1 +

2! 3! 4! + + + …∞ 22 33 4 4

[Ans.: Convergent]

[Ans.: Convergent] [Ans.: Convergent]

1.62

5.

Chapter 1

2 3 4 + + + 3! 4 ! 5!

∞

[Ans.: Convergent]

3 32 33 34 + + + + 2 ! 3! 4 ! 5!

6. 1 +

7.

Sequences and Series

[Ans.: Convergent]

n2

∑3 n =1

n

[Ans.: Convergent] ∞

8.

2

∑3 n =1 ∞

9.

n −1

[Ans.: Convergent]

+1

n

1

∑ n! n =1

∞

10.

[Ans.: Convergent]

n (n + 1) n! n =1

∑

2

2

[Ans.: Convergent]

∞

3n + 4 n 11. ∑ n n n =1 4 + 5

[Ans.: Divergent]

∞

xn 12. ∑ n 2 , x > 0 n =1 3 ⋅ n ∞

13.

3 − 2 n −1 ⋅ x , x>0 n +1 n =1

∑3 ∞

14.

[Ans.: Convergent for x < 1, divergent for x > 3]

xn

∑ (2 )! n =1 ∞

15.

[Ans.: Convergent for x < 3, divergent for x > 3]

n

∑

n =1

n

[Ans.: Convergent] n +1 n ⋅x , x > 0 n3 + 1 [Ans.: Convergent for x < 1, divergent for x > 1]

16. x + 2 x 2 + 3x 3 + 4 x 4 + … ∞ [Ans.: Convergent for x < 1, divergent for x > 1] 2

17. 1 +

3

x x x xn + + +…+ 2 + …∞ 2 5 10 n +1 [Ans.: Convergent for x < 1, divergent for x > 1]

1.9

18.

Cauchy’s Root Test

1.63

x x2 x3 + + + …∞ 1⋅ 3 3 ⋅ 5 5 ⋅ 7 [Ans.: Convergent for x < 1, divergent for x > 1]

19. x +

20.

3 2 8 3 15 4 n2 − 1 n x + x + x +… + 2 x + …∞ 5 10 17 n +1 [Ans.: Convergent for x < 1, divergent for x > 1]

x 2 3

+

x2 3 4

+

x3 4 5

+ …∞ [Ans.: Convergent for x < 1, divergent for x > 1]

1.9

CAUCHY’S ROOT TEST 1

If Sun is a positive term series and if lim (un ) n = l then n Æ•

(i) Sun is convergent if l < 1. (ii) Sun is divergent if l > 1. Proof 1

Case I If lim (un ) n = l < 1. n Æ•

1

Consider a number l < r < 1 such that (un ) n < r for all n > m un < r n for all n > m The geometric series

... (1)

Sr n = r + r 2 + r 3 + …∞

Sn = r + r 2 + r 3 + … + r n r (1 - r n ) 1- r r (1 - r n ) lim Sn = lim n Æ• n Æ• 1 - r r = , which is finite 1- r =

Hence, the series Sr n is convergent. From Eq. (1),

un < r n for all n > m

˘ È∵r < 1 ˙ Í n r = 0˙ ÍÎ\ xlim Æ• ˚

1.64

Chapter 1

Sequences and Series

Sun < Sr n Since Sr n is convergent, Sun is also convergent. 1

Case II: If lim ( ) n Æ•

1. 1

(un ) n > 1 for all n > m

… (2)

Neglecting the first m terms, 1

1

1

1

S(

+ …∞ >

[Using Eq. (2)]

1… ∞ 1

1

1

1 m n

S >

erms

n

lim Sn > lim n Æ •

n Æ•

n Æ•

•

The series

Â

un is divergent. The nature of a series remains unchanged if we

n = m +1

•

neglect a finite number of terms in the beginning. Hence, the series

 un

is divergent.

n 1 1

Note: If lim( ) n→∞

1, the root test fails, i.e., no conclusion can be drawn about the

convergence or divergence of the series.

Example 1 •

Test the convergence of the series

1

Â

n 1 (log n)

n

Solution un =

Let

1

1 (log n)n

1 log n

( )n 1

1 log n 1 [ log • Æ •]

lim ( ) n = lim

n Æ•

n Æ•

=

1.9

Hence, by Cauchy’s root test, the series is convergent.

Example 2 a n +1 . ∑ n n =1 n ∞

Test the convergence of the series

Solution un =

Let

a n +1 nn 1+

1 n

( ) (a ( n) = (u n

1 n

1 1 n

a an lim(un ) = lim n →∞ n →∞ n = 0 1 and is divergent if p £ 1. 1 2 3 n=1 n [Summer 2015] Solution Let

un = f ( x) = •

Ú1

1 np 1

= f ( n)

xp

f ( x )dx = Ú

• 1

1 xp

dx

1.76

Chapter 1

Sequences and Series

x - p +1 = lim m Æ• - p + 1

m

1

Êm 1 ˆ = lim Á m Æ• Ë 1 - p 1 - p ˜¯ 1- p

1 , 1- p = •, =-

If

p = 1,

•

Ú1

f ( x )d dxx = Ú

• 1

p >1 p 1 and is infinite if p £ 1.

Hence, by Cauchy’s integral test, the series is convergent if p > 1 and is divergent if p £ 1.

EXERCISE 1.5 Test the convergence of the following series: ∞

1.

∑

n =1

1 n [Ans.: Divergent]

∞

1 2. ∑ 2 n =1 n + 1 ∞

3.

∑ ne

[Ans.: Convergent] − n2

n =1 ∞

[Ans.: Convergent]

1 4. ∑ 2 n =1 n(log n) [Ans.: Convergent]

1.11 Alternating Series

1.77

1.11 ALTERNATING SERIES An infinite series with alternate positive and negative terms is called an alternating series.

Leibnitz’s Test for Alternating Series An alternating series

•

•

n= n =1

n =1

1)n -1vn =  un is convergent if  (--1)

(i) each term is numerically less than its preceding term, i.e. un +1 < un or un > unn+1 (ii) lim un = 0 n Æ•

Example 1 Test the convergence of the series 1 − Solution Let

un = ( -1)n -1 un =

1 1 1 + − +… . 2 3 4

1 n

1

n The given series is an alternating series. 1 1 un - un +1 = (i) n n +1

=

n +1 - n n n +1

>0

for all n ΠN

\ un > un +1 (ii)

lim un = lim

nnƕ ƕ

1

n Æ•

n =0 Hence, by Leibnitz’s test, the series is convergent.

Example 2 Test the convergence of the series 1 Solution Let

un =

1 2 2

+

( -1)n -1 n n

1 3 3

-

1 4 4

+ +.

1.78

Chapter 1

Sequences and Series

1

un =

n n

The given series is an alternating series. 1 1 (i) un - un +11 = n n (n + 1) n + 1 (n + 1) n + 1 - n n > 0 for all n Œ N ( n n ) ÈÎ(n + 1) ( n + 1) ˘˚ \ un > un +1 =

1

lim un = lim

(ii)

n Æ•

n Æ•

n n =0 Hence, by Leibnitz’s test, the series is convergent.

Example 3 Test the convergence of the series

1 2

1

-

1 2

2

+

Solution un = ( -1)n -1 ◊

Let

un =

1 2

3

-

1 42

+ .

1 n2

1 n2

The given series is an alternating series. (i) un - un +11 =

1

-

1

(n + 1)2 2n + 1 2n = 2 > 0 for all n ΠN n (n +1 + 1)2

\

n

2

un > un +1

(ii) lim un = lim n Æ•

n Æ•

1 n2

=0 Hence, by Leibnitz’s test, the series is convergent.

Example 4 Test the convergence of the series

1 1p

-

1 2p

+

1 3p

-

1 4p

+ º.

1.11 Alternating Series

Solution Let

1

un = ( -1)n un =

np

1 np

The given series is an alternating series. Case I: If p > 0, 1 1 (i) un - un +1 = p n (n + 1) p =

(n + 1) p - n p n p (n + 1) p

>0

[∵ p > 0]

\ un > un +1 (ii) lim un = lim nnƕ ƕ

n Æ•

=0

1 np [∵ p > 0]

Hence, by Leibnitz’s test, the series is convergent if p > 0. Case II: If p < 0 In this case the conditions (i) and (ii) of the Leibnitz’s test are not satisfied. Hence, the given series is not convergent if p < 0.

Example 5 Test the convergence of the series 1 -

1 1 1 1 + - + -. 2 4 8 16

Solution un = ( -1)n -1

Let

un =

1 n -1

2 The given series is an alternating series. 1 1 (i) un - un +1 = nn-1 - n 2 2 1 Ê 1ˆ = n -1 Á 1 - ˜ 2 Ë 2¯

1 1 ◊ 2 n -1 2 1 = n > 0 for all n Œ N 2 =

1 2 n -1

1.79

1.80

Chapter 1

\

Sequences and Series

un > un +1

(ii) lim un = lim n Æ•

n Æ•

1 2

n -1

=0

Hence, by Leibnitz’s test, the series is convergent.

Example 6 Test the convergence of the series

1 2 3 4 - + - +. + 2 5 10 17

Solution un = ( -1)n -1

Let

un =

n n +1 2

n n2 + 1

The given series is an alternating series. (i) un - un +11 = = =

n n +1 2

-

n +1 (n + 1)2 + 1

n(n2 + 2 n + 2) - (n + 1)( )(n2 + 1) (n2 + 1)(n2 + 2 n + 2) n2 + n - 1 (n2 + 1)(n2 + 2 n + 2)

> 0 for all n ΠN

\ un > un +1

(ii) lim un = lim

n

n2 + 1 1 = lim 1 n Æ• n+ n =0 Hence, by Leibnitz’s test, the series is convergent. n Æ•

n Æ•

Example 7 ((-1)n+1 . Â n£1 log( n + 1) •

Test the convergence of the series Solution Let

un =

( -1)n +1 log(n + 1)

[Summer 2014]

1.11 Alternating Series

1 log(n + 1)

un =

The given series is an alternating series. (i) un - un +11 = =

1 1 log(n + 1) log (n + 2)

log(n + 2) - llog( og(n + 1) >0 log(n + 1) ◊ llog og (n + 2)

\ un > un +1 1 n Æ• log( n + 1)

(ii) lim un = lim n Æ•

=0 Hence, by Leibnitz’s test, the series is convergent.

Example 8 •

Test the convergence of the series Solution

Â

n =1 n

un =

Let

((-1)n -1

un =

2

(n + 1)

.

( -1)n -1 n2 (n + 1) 1 n (n + 1) 2

The given series is an alternating series. (i) un - un +11 = =

1 n (n + 1) 2

-

1 (n + 1) (n + 2) 2

(n + 1)(n + 2) - n2

=

n2 (n + 1)2 ((n n + 2)

3 +2 3n n2 (n + 1)2 (n + 2)

> 0 for all n ΠN

\ un > un +1

(ii) lim un = lim n Æ•

n Æ•

1 n2 (n + 1)

=0 Hence, by Leibnitz’s test, the series is convergent.

1.81

1.82

Chapter 1

Sequences and Series

Example 9

x2

Test the convergence of the series x

x3

x4

1) .

Solution 1)n

un

Let

1

xn n

xn n

un

The given series is an alternating series. (i) un

n +1

=

x n x n +1 n n +1

xn n

=

xn

=

\ (ii)

[

>0

un +1

lim

n Æ•

n

] 1)

n

u

lim

nx

xn n

=0 Hence, by Leibnitz’s test, the series is convergent.

Example 10 Test the convergence of the series log

1

2 + log 3

log

log

4 + 5

log

4 + 5

Solution log

1

log

- log

Let

2 3 + log 3

2 3 + log 1 2

un = - )n log un

log

n +1 n

n +1 n

log

4

+ log

5 4



1.11 Alternating Series

1.83

The given series is an alternating series. (i) un - un +11 = logg

n +1 n+2 È n +1 n + 2 ˘ - log > 0 Í∵ > for all n Œ N ˙ n n +1 n n +1 Î ˚

\ un > un +1

Ê n + 1ˆ (ii) lim un = lim log Á Ë n ˜¯ n Æ• n Æ•

Ê 1ˆ = lim log Á 1 + ˜ Ë n Æ• n¯ = log 1 =0 Hence, by Leibnitz’s test, the series is convergent.

Example 11 Test the convergence of the series

1 2 3 4 - + - +. + 2 3 4 5

Solution un = ( −1) n −1 ⋅

Let

un =

n n +1

n n +1

The given series is an alternating series. n n +1 (i) un - un +11 = n +1 n + 2 n2 + 2n - n2 - 2n - 1 (n + 1)(n + 2) 1 =1 8 = lim

•

By D’Alembert’s ratio test,

 un

is convergent. Thus, the series is absolutely conver-

n=1

gent and hence is convergent.

Example 2 Test the series for absolute or conditional convergence 2 3 4 1- + 2 + 3 +. 3 3 3 Solution un = ( -1)n -1 ◊

Let •

 un

= 1+

n =1

un = un +1 = lim

n Æ•

un un +1

n 3n -1

2 3 4 + + +… 3 32 33

n n -1

3 n +1 3n

= lim

n Æ•

n

3

3

= lim

n Æ•

◊ n -1

1+

= 3>1

1 n

3n n +1

1.86

Chapter 1

Sequences and Series

•

By D’Alembert’s ratio test, convergent.

 un

is convergent and hence, the series is absolutely

n =1

Example 3 •

Test the series for absolute or conditional convergence

Â

(-1)n

. n + n +1 [Winter 2016]

n =1

Solution (-1)n n + n +1

un =

Let

1

un =

n + 1+n

The given series is an alternating series. 1 1 (i) un - un +1 = n + 1+ n n +1 + n + 2 = = \ (ii)

( n + 1 + n ) ( 1 + n + n + 2) n+2 - n ( n + 1 + n ) ( 1 + n + n + 2)

un > un +1

lim un = lim

n Æ•

By Leibnitz’s test, un = =

Let

n +1 + n + 2 - n - n +1

vn =

1

=0

n Æ•

n + 1+ n

 un

is convergent. 1

n + 1+ n 1 Ê 1ˆ n Á1 + 1 + ˜ n¯ Ë 1 n

>0

for all n ΠN

1.12 Absolute Convergence of a Series

lim

n Æ•

un

= lim

n Æ•

vn

1 Ê 1ˆ Á1 + 1 + n ˜ Ë ¯ =6

and

 vn = Â

1

 un

[finite and non-zero]

1 is divergent as p = . 2

1 n2

By comparison test, The series

1.87

 un

is also divergent.

is convergent and

 un

is divergent.

Hence, the series is conditionally convergent.

Example 4 Determine absolute or conditional convergence of the series •

n2

n =1

n3 + 1

 ((--1)n ◊

[Winter 2013; Summer 2017]

.

Solution un = ( -1)n ◊

Let

un =

n2 n3 + 1

n2

n +1 1 = 1ˆ Ê n Á1 + 3 ˜ Ë n ¯

vn =

Let lim

n Æ•

un vn

3

1 n

= lim

1

1 n3 = 1 [[finite finite and non-zero] n Æ•

1+

1 is divergent as p = 1. n By comparison test, Â un is also divergent.

and Svn = Â

Hence, Sun is not absolutely convergent. To check the conditional convergence, applying Leibnitz’s test, (i) un - un +11 =

n2 n +1 3

-

(n +1 + 1)2 (n + +11)3 + 1

1.88

Chapter 1

Sequences and Series

n2 (n3 + 3n2 + 3n + 2) - (n3 + 1)(n2 + 2 n + 1)

=

( n3 + +1 1)[(n + 1)3 + 1] n 4 + 2 n3 + n 2 - 2 2n n -1

=

= =

( n3 + +1 1)[( 1 )[(n + 1)3 +1 + 1]

n 4 + n2 (2 n + 1) -1 1(2 n + 1) (n3 + 1)[(n + 1)3 + 1] n 4 + (2 n + 1)(n2 - 1) (n3 + 1)[[((n + 1 1))3 + 1]

> 0 for all n ΠN

un > un +1

(ii) lim un = lim n Æ•

n Æ•

= lim

n Æ•

=0 By Leibnitz’s test,

n2 n3 + 1

1 1ˆ Ê n Á1 + 3 ˜ Ë n ¯

Su

n

is convergent.

The series Sun is convergent and

 un

is divergent.

Hence, the series is conditionally convergent.

Example 5 Test the series for absolute or conditional convergence 2 3 1 4 1 5 1 - ◊ + ◊ - ◊ +º. 3 4 2 5 3 6 4 Solution Ê n +1 1ˆ un = ( -1)n -1 Á ◊ Ë n + 2 n ˜¯

Let •

Â

un =

2 3 1 4 1 5 1 + ◊ + ◊ + ◊ + ◊◊◊. 3 4 2 5 3 6 4

| un | =

n +1 1 ⋅ n+2 n

n=1

vn =

Let lim

n Æ•

un vn

1 n

= lim

n Æ•

n +1 n+2

1.12 Absolute Convergence of a Series

1.89

1 n = lim 2 n Æ• 1+ n = 1 [ffinite inite and non-zero] 1+

1 is divergent as p = 1. n By comparison test, Σ un is also divergent. Hence, the series is not absolutely convergent. To check the conditional convergence, applying Leibnitz’s test, and Σvn = ∑

(i) un - un +1 = =

n +1 n+2 nn((n + 2) (n + 1)(n + 3)

n 2 + 3n + 3 >0 nn((n + 1)(n + 2)(n + 3)

for all n ΠN

un > un+ n 1 n +1 n Æ• n( n + 2) 1 1+ n = lim n Æ• Ê 2ˆ n Á 11+ + ˜ Ë n¯

(ii) lim un = lim nnƕ ƕ

=0 By Leibnitz’s test, Sun is convergent. The series Sun is convergent and Σ un is divergent. Hence, the series is conditionally convergent.

Example 6 Test the convergence of the series x −

x3 x5 + − − , x > 0. 3 5 [Summer 2016]

Solution Let

un = ( -1)n -1 un =

n -1 x 2 n2n - 1

un +1 =

n +1 x 2 n+ 2n + 1

n -1 x 2 n2n - 1

1.90

Chapter 1

Sequences and Series

un un +1

=

n -1 x 2 n2n + 1 ◊ 2 n+ 2 n - 1 x n +1

1 n◊ 1 = 1 x2 2n 1ˆ Ê 2+ Á n˜◊ 1 = lim Á 1 ˜ x2 n Æ• Á2- ˜ Ë n¯ 2+

lim

n Æ•

un un +1

=

 un

1 x2 1

> 1 or x2 < 1 or x < 1 [∵ x > 0] x2 Thus, the given series is absolutely convergent and hence, is convergent for x < 1. If x2 = 1 or x = 1 [∵ x > 0] ( -1)n -1 un = 2n - 1 1 un = 2n - 1 The given series is an alternating series.

By D’Alembert’s ratio test,

is convergent if

1 1 2n - 1 2n + 1 2 = 2 > 0 ffor or all n ΠN 4n - 1

(i) un - un +11 =

1 n Æ• n Æ• 2 n - 1 =0 By Leibnitz’s test, the series is convergent for x = 1 Hence, the series is convergent for x £ 1.

(ii) lim un = lim

EXERCISE 1.7 Test the following series for absolute or conditional convergence: 1. 1 −

1 1 1 1 1 + − + − +… 2 3 4 5 6

[Ans.: Conditionally convergent] 2. 1 +

1 1 1 1 1 − − + + −… 2 2 32 4 2 52 6 2

[Ans.: Absolutely convergent]

1.13

1

3.

2

1

+

3

−

1 4

Uniform Convergence of a Series

1.91

+…

[Ans.: Conditionally convergent] sin 4. 1

1.13

s

s

x

−…

[Ans.: Absolutely convergent]

UNIFORM CONVERGENCE OF A SERIES ∞

The series

∑

of real-valued functions defined in the interval (a, b) is said to

n

n 1

converge uniformly to a function S (x) if for a given ∈> 0, there exists a number m independent of x such that for every x ( ),

|

< ∈ for all n > m

( )

S

where,

) ∞

∑

Weierstrass’s M-Test The series

is said to converge uniformly in an

n

n =1

∞

interval (a, b), if there exists a convergent series

∑M

n

of positive constants such that

n 1

M n for all x (

)

∞

Proof Let

∑M

n

is convergent, then for a given ∈> 0, there exists a number m such

n =1

that

< ∈ for all n > m,

… ∞ and C

where

Mn

< ∈ for all n > m

then

< ∈ for all n > m

( Now, |

M

M n for all x ∈ (

[∵ Mn is positive constant]

) ≤ Mn

+…

< Πfor all n > m \ where

< ∈ for all n > m Sn (x) = u1(x) + u2 (x) + … + un (x) ∞

Since m does not depend on x, the series ∑ n 1 (a, b).

n

converges uniformly in the interval

1.92

Chapter 1

Example 1 Test the series

Sequences and Series

∞

∑n n =1

4

1 for uniform convergence. + n3 x 2

Solution 1 n 4 + n3 x 2 1 | un ( x ) | = 4 n + n3 x 2 un ( x ) =

Let


1. n =1 n

=∑

Hence, by M-test, the series is uniformly convergent for all real values of x.

Example 2

∞

sin( x 2 + n 2 x) Test the series ∑ for uniform convergence . n(n 2 + 2) n( n =1 Solution Let

un ( x ) =

sin( x 2 + n 2 x) n(n 2 + 2)

| un ( x ) | =

sin( x 2 + n 2 x) n ( n 2 + 2)

=

∞

| sin(( x 2 + n 2 x) | n ( n 2 + 2)

≤

1 n + 2n

for all x ∈ R


1. n =1 n

∑ Mn = ∑ n= =11

Hence, by M-test, the series is uniformly convergent for all real values of x.

1.13

Example 3 Test the series sin x −

Uniform Convergence of a Series

1.93

sin 2 x sin 3 x sin 4 x + − + … . for uniform 2 2 3 3 4 4

convergence. Solution un ( x) = ( −1) n −1

Let

sin nx n n

sin nx n n

| un ( x ) | =

1

≤

n

3 2

for all x ∈ R

∵ −1 ≤ sin q ≤ 1  | sin q | ≤ 1 

1

Mn =

3

n2 ∞

∑M nn= =1

∞

n

=∑ n =1

1 3 2

is convergent as p =

3 > 1. 2

n Hence, by M-test, the series is uniformly convergent for all real values of x.

Example 4 ∞

Show that if 0 < r < 1, the series

n

cos n 2 x is uniformly convergent.

n =1

Solution Let

∑r

un ( x ) = r n cos n2 x un ( x ) = r n cos n2 x £|| r n | for all x Œ R £ = rn, 0 < r < 1

cos q £ 1˘ È∵- 1 £ cos Í cos q £ 1 ˙ Î ˚

Mn = r n •

•

n= n =1

n =1

 Mn =  r n = r + r 2 + r 3 + … which is convergent being a geometric series with 0 < r < 1. Hence, by M-test, the series is uniformly convergent for all real values of x.

1.94

Chapter 1

Sequences and Series

EXERCISE 1.8 Test the following series for uniform convergence: ∞

1.

sin(x 2 + nx) ; for all real x n(n + 2) n( n =1

∑ ∞

2.

1 ; for all real x and p > 1 ∑ p q 2 n =1 n + n x

[[Ans.: Uniformly convergent]

[Ans.: Uniformly convergent] 3.

sin x sin2 x sin3x sin 4 x + + + +… 12 22 32 42

[Ans.: Uniformly convergent] 4. Show that if 0 < r < 1, then the series ∞

∑r

n

sin a n x is uniformly convergent for all real values of x.

n =1

5. Show that

1 1 1 1 − + − +… 2 2 2 1+ x 2+ x 3+ x 4 + x2

converges uniformly in the interval x ≥ 0 but not absolutely.

1.14

POWER SERIES ∞

A power series is an infinite series of the form

∑a x n

n

= a0 + a1 ( x − c) + ..

n =1

a2 ( x − c) 2 + a3 ( x − c)3 +  where an represents the coefficient of the nth term, c is a constant and x varies around c. When c = 0, the series becomes ∞

∑a x n

n

= a0 + a1 x + a2 x 2 + a3 x 3 + 

n=0

1.14.1

Interval and Radius of Convergence

A power series will converge only for certain values of x. An interval (−R, R) in which a power series converges is called the interval of convergence. The number R is called the radius of convergence, e.g., if a power series converges for all the values of x, then interval of convergence will be (−•, •) and the radius of convergence will be •.

1.14.2 Test for Convergence Since a power series may be positive, alternating or mixed series, the concept of absolute convergence is used to test the convergence of a power series. Applying D’Alembert’s ratio test,

1.13

Power Series

1.95

un = an x n un +1 = an +1 x n +1 lim

n Æ•

un an x n = lim un +1 nƕ an +1 x n +1 =

If

a 1 lim n Æ• • an +1 x nÆ

lim

an = l, an +1

lim

un 1 l = ◊l = un +1 x x

n Æ•

n Æ•

By D’ Alembert’s ratio test, the series is absolutely convergent and hence is convergent l > 1, i.e., | x |< l, - l < x < l. If x Here, interval of convergence of the series is (−l, l ) and the radius of convergence is l.

Example 1 ∞

xn Obtain the range of convergence of ∑ n , x > 0. n =1 2 Solution un =

Let

un +1 =

xn 2n x n +1 2 n +1

un x n 2 n +1 = lim n ◊ n +1 n Æ• un +1 n Æ• 2 x 2 = x By D’Alembert’s ratio test, the series is 2 (i) convergent if > 1 or x < 2 x 2 (ii) divergent if < 1 or x > 2 x lim

The test fails if Then

2 = 1, or x = 2. x un =

2n 2n

=1

1.96

Chapter 1

Sequences and Series

•

 un = 1 + 1 + 1 +  • n=1

which is a divergent series. Hence, the series is convergent for 0 < x < 2 and the range of convergence is 0 < x < 2.

Example 2 ∞

2n x n Determine the interval of convergence for the series ∑ and also, n = 0 n! their behaviour at each end points.

Solution un =

Let

un +1 =

2n x n n! 2 n +1 x n +1 (n + 1)!

un 2 n x n (n + 1)! = ◊ un +1 n ! 2 n +1 x n +1 n +1 = 2x

lim

n Æ•

un n +1 = lim un +1 nƕ 2 x

= • >1 Hence, By D’Alembert’s ratio test, the series is convergent for all values of x i.e. – • < x < • and interval of convergence is (– •, •).

Example 3 ∞

Obtain the range of convergence of

xn , x > 0, a > 0. ∑ n n =1 a +

Solution un =

Let

un +1 =

xn a+ n

x n +1 a + n +1

un xn a + n +1 = lim ◊ n Æ• un +1 n Æ• a + n x n +1 lim

1.13

By D’Alembert’s ratio test, the series is 1 (i) convergent if > 1 or x < 1 x 1 (ii) divergent if < 1 or x > 1 x The test fails if x = 1.

=

vn =

Let

1.97

a 1 + 1+ n 1 n = lim ◊ a n Æ• x +1 n 1 = x

un =

Then

Power Series

1 a+ n

1  1  n  a + 1   n 

1 n

u 1 lim n = lim n Æ• vn n Æ• a +1 n = 1 [ffinite inite and non-zero] and Σvn = ∑

1 1 2

is divergent as p =

1 < 1. 2

n By comparison test, Σun is also divergent for x = 1. Hence, the series is convergent for 0 < x < 1 and the range of convergence is 0 < x < 1.

Example 4

∞

( x + 1) n Obtain the range of convergence of ∑ n . n =1 3 ⋅ n Solution Let

un = un +1 =

( x + 1)n 3n ◊ n ( x + 1)n +1 n +1 +1 (n + 1) 3n+

1.98

Chapter 1

Sequences and Series

n +1 +1 un ( x + 1)n 3n+ (n + 1) = n ◊ un +1 3 ◊ n ( x + 1)n +1 3(n + 1) = ( x + 1)n

Ê 1ˆ 3 Á1 + ˜ Ë n¯ = x +1  1 3 1 +   n un lim m = lim n →∞ u n →∞ x +1 n +1 3 x +1

=

The series is convergent if 3 >1 x +1 3 > x +1

x+ +11 < 3 −3 < ( x + 1) < 3 −4 < x < 2

At x = 2, un = ∴

∞

∑u n =1 =1

1 n

∞

n

1 n =1 n

=∑

is divergent as p = 1. ( −1) n n 1 un = n un =

At x = – 4,

The given series is an alternating series. 1 1 − n n +1 1 = >0 n(n +1 + 1)

(i) un − un +1 =

∴ u n > u n +1 m un = lim (ii) lim n →∞

n →∞

=0

1 n

for all n ΠN

1.13

Power Series

1.99

By Leibnitz’s test, the series is convergent at x = – 4. Hence, the series is convergent for – 4 £ x < 2 and the range of convergence is – 4 £ x < 2.

Example 5

∞

Obtain the range of convergence of

∑ n+ n =1

Solution un =

Let

u n +1 =

xn 1 + n2

.

xn n + 1 + n2

x n +1 (n + +11) + 1 + (n + 1) 2

(n + +11) + 1 + (n + 1) 2 un xn = ⋅ u n +1 n + 1 + n 2 x n +1

1  1  1 + 1 +  1 +  + n n2  n  =   1 + 1 x 1 + 2  n

lim m

n →∞

un u n +1

1  1  1 + 1 +  1 +  + n n2  n  = lim n →∞   1 + 1 x 1 + 2  n =

The series is convergent if

At x = 1,

1 x

1 >1 x x 0

for all n ŒN

∴ u n > u n +1

(ii) lim m un = lim n →∞

n →∞

1 n + 1 + n2

=0

Thus, by Leibnitz’s test, the series is convergent if x = –1. Hence, the series is convergent for – 1 £ x < 1 and the range of convergence is – 1 £ x < 1.

Example 6 •

For the series convergence.

 (-1)

n =1

n -1

x2n - 1 , find the radius and interval of 2n - 1 [Winter 2016]

1.13

Power Series

1.101

Solution Let

un =

(-1)n -1 x 2 n - 1 2n - 1

un + 1 =

(-1)n x 2 n + 2 - 1 2n + 2 - 1

=

(-1)n x 2 n + 1 2n + 1

un (-1)n - 1 x 2 n -1 2n + 1 = ◊ 2n - 1 un + 1 (-1)n x 2 n + 1 =-

lim

n Æ•

(2 n + 1) 1 ◊ (2 n - 1) x 2

un (2 n + 1) 1 = lim ◊ n Æ• (2 n - 1) x 2 un + 1 =

1 x2

By D’Alembert’s ratio test, the series is convergent if i.e.,

– 1 < x < 1 and divergent for x > 1.

At x = 1, un =

(-1)n - 1 (1)2 n - 1 2n - 1

=1= un =

1 1 1 + - + 3 5 7

(-1)n - 1 2n - 1 1 2n - 1

The given series is an alternating series. (i)

un - un + 1 =

1 1 2n - 1 2n + 1

=

2n + 1 - 2n + 1 (2 n - 1) (2 n + 1)

1 2 2 > 1 or 1 > x or x < 1 x2

1.102

Chapter 1

Sequences and Series

=

2 >0 (2 n - 1) (2 n + 1)

for all n ΠN

un > un +1 lim un = lim

(ii)

n Æ•

n Æ•

By Leibnitz’s test,

 un

1 =0 2n - 1 is convergent.

At x = –1, un = = un =

(-1)n - 1 (-1)2 n - 1 2n - 1 (-1)3n - 2 2n - 1 1 2n - 1

The given series an alternating series. Hence, by Leibnitz’s test, Â un is convergent. Thus, for the interval –1 £ x £ 1, given series is convergent. Since condition for convergence is | x2 | < 1, radius of convergence = 1. •

Since

 un

is convergent, the series is absolutely convergent for –1 £ x £ 1.

n =1

Example 7 •

Find the interval of convergence of the series

Â

( -3)n x n

. n +1 [Winter 2013; Summer 2016] n= 0

Solution un = un +1 =

(-3)n x n n +1 (-3)n +1 x n +1 n+2

1.13

Power Series

1.103

un (-3 -3)n x n n+2 = ◊ un +1 -3)n +1 x n +1 n + 1 (-3 =

n+2 (-33) x n + 1 1+

=

2 n

(-33) x 1 +

u lim n = lim n Æ• un +1 n Æ•

1 n

1+

2 n

( -3) x 1 +

1 n

1 -3 x 1 = 3x =

1 1 > 1 or 3 x < 1 or x < or By D’Alembert’s ratio test, the series is convergent if 3x 3 1 1 - 1 x +1 2 > x +1

x+ +11 < 2 −2 < ( x + 1) < 2 −3 < x < 1

The series is convergent in the interval (– 3, 1). At x = –3, un = ( −1) n ∞

∑u n=1 n

n((− −3 + 1) n 2n

∞

n

= ∑n n =1 n=

which is a divergent series. At x = 1, n((11 + 1) n 2n n = ( −1) n

un = ( −1) n un = n lim un ≠ 0

n →∞

By Leibnitz’s test, the series is not convergent at x = 1. Hence, the series is not convergent at each end point and the interval of convergence is (– 3, 1).

1.106

Chapter 1

Sequences and Series

Example 9 •

For the series

( -1)n ( x + 2)n , find the radius and interval of  n n =1

convergence. For what values of x does the series converge absolutely, conditionally? [Winter 2015] Solution un =

Let

un +1 =

(-11)n ( x + 2)n n (-11)nn++1 ( x + 2)n +1 n +1

un (-11)n ( x + 2)n ((n + 1) = ◊ n+ n + un +1 n (-11) 1 ((xx + 2)n +1 1 Ê 1ˆ = - Á1 + ˜ ◊ Ë n ¯ ( x + 2) lim

n Æ•

un 1 Ê 1ˆ = lim Á 1 + ˜ ◊ un +1 nÆ• Ë n ¯ ((x + 2) =

1 x+2

The series is convergent if 1 >1 x+2 1> x+2

x+2 1, u lim m n = lim n →∞ u n →∞ n +1

1 x2n 1   x 1 + 2 n   x  x2 +

= x > 1 ∵ lim lim x 2 n → ∞   n→∞ 

Thus, the series is convergent for x > 1, i.e. x > 1 and x < – 1 At x = 1, 1 2 ∞ 1 1 1 un = + + +  ∞ ∑ 2 2 2 n =1 un =

which is a divergent series. At x = – 1,

( −1) n 2 1 un = 2 1 lim un = ≠ 0 n →∞ 2 un =

Thus, by Leibnitz’s test, the series is not convergent at x = – 1. Hence, the series is convergent for x > 1 and range of convergence is x > 1.

EXERCISE 1.9 Obtain the range of convergence of the following series: 1 x + 2 x 2 + 3 x 3 +  + nx n +  1. 1+ 1 x x2 x3 xn 2. + + + ++ + 2 3 4 5 n+2

[Ans.: −1< x < 1]

[Ans.: −1 < x < 1] ∞

(−1)n x n 3. ∑ n =1 (n + 1) ∞

4.

∑ n=0 ∞

5.

(x + 2) n +1

∑ (−1)n n=0

xn log(n + 1)

[Ans.: |x| ≤ 1] [Ans.: −3 ≤ x ≤ −1] [Ans.: |x| < 1]

Points to Remember

∞

6.

1 3   Ans.: 2 < x < 2   

∑ (−2) (n + 1))((x − 1) n

1.109

n

n=0 ∞

7.

∑ n !(x − 1)

n

[Ans.: x = 1]

n =1 ∞

8.

(n !!)2

∑ (2n))! x

n

[Ans.: |x| < 4]

n =1 ∞

(−2)n (2 x + 1)n 9. ∑ n2 n =1 ∞

10.

∑

n =1

3 1   Ans.: − 4 ≤ x ≤ − 4   

(−1)n x 2 n 3

[Ans.: −1 ≤ x ≤ 1]

(2n) 2

POINTS TO REMEMBER • Sequence A sequence {un} is said to be convergent, divergent or oscillatory according as lim un is finite, infinite or not unique respectively. n→∞

• Series The infinite series Sun is said to be convergent, divergent or oscillatory according as lim Sn is finite, infinite or not unique respectively. n→∞

If a positive term series Sun is convergent then lim un = 0 but n→∞

converse is not true i.e., if lim un = 0, the series may converge or n→∞

diverge. If lim un π 0, the series is not convergent. n→∞

• Comparison Test If Sun and Svn are series of positive terms such that nlim →∞

un = l vn

(finite and non-zero) then both series converge or diverge together. • D’Alembert’s Ratio Test

un = l then un +1 if l > 1.

If Sun is a positive term series and nlim →∞ (i) Sun is convergent

1.110

Chapter 1

Sequences and Series

(ii) Sun is divergent (iii) The test fails

if l < 1. if l = 1.

• Cauchy’s Root Test

1

If Sun is a positive term series and if lim(un) n = l then (i) Sun is convergent if l < 1. (ii) Sun is divergent if l > 1.

n→∞

This test is preferred when un contains nth powers of itself. • Cauchy’s Integral Test If Sun = S f (n) is a positive term series where f (n) decreases as n in∞

∫

creases and let

1

f (x) dx = I then

(i) Sun is convergent if I is finite. (ii) Sun is divergent if I is infinite. This test is preferred when evaluation of the integral of f (x) is easy. • Leibnitz’s Test

∞

An alternating series

∑ (− 1)

n −1

un is convergent if

n =1

(i) each term is numerically less than its preceding term, i.e, un +1 < un or un > un +1 (ii) lim un = 0 n→∞

• Absolute Convergence ∞

∑u

The series

n

with both positive and negative terms (not neces-

n =1

sarily alternative) is called absolutely convergent if the correspond∞

ing series

∑ ||u | with all positive terms is convergent. n

n =1

• Conditional Convergence ∞

If the series

∑u n =1

∞

n

is convergent and

∑ ||u | is n

n =1

∞

series

∑u n =1

n

is called conditionally convergent.

divergent then the

Multiple Choice Questions

1.111

• Uniform Convergence (Weierstrass’s M-Test) ∞

The series

∑ u ( x) n

is said to converge uniformally in an interval

n =1

(a, b), if there exists a convergent series of positive constants such that |un(x)| £ Mn for all x Œ (a, b).

MULTIPLE CHOICE QUESTIONS Choose the correct alternative in the following questions: 1 is x

1. The value of lim x sin x Æ0

(b) p

(a) 1

(c) 0

•

2. The sum of the series

Â

n =1

(a) 0

(b)

[Summer 2014] (d) •

1 is 2n

3 4

[Winter 2015] (c) 1

(d) 2

•

3. The series

( -1)n is n =1 n

Â

[Winter 2015]

(a) divergent (c) conditionally convergent

(b) absolutely convergent (d) nothing can be said

1

4. The series

 n(5 n - 1)

is

(a) convergent (c) oscillates finitely •

5. The series

Â

n=2

(b) divergent (d) oscillates infinitely

1 is n(log n)

(a) convergent (c) oscillates finitely 1 6. The series  is divergent if np (a) p > 1 (b) p £ 1

(b) divergent (d) oscillates infinitely

(c) p = 1

•

un +1 > 1, then  un is x Æ 0 un n =1 (a) convergent (c) may or may not be convergent

7. If lim

(b) divergent (d) oscillatory

(d) p = 0

1.112

Chapter 1

Sequences and Series

2 3 8. The series a + ar + ar + ar +  oscillates finitely if

(a) |r| < 1

(b) r > 1

3 9 27 81 + + + +  is 4 8 16 32 (a) convergent (c) oscillates finitely

(c) r = 1

(d) r £ –1

9. The series

•

10. The series

Â

n =1

1

(b) divergent (d) oscillates infinitely

is

n 5n

(a) convergent (b) divergent (c) oscillates finitely (d) oscillates infinitely 2 3 11. The series a + ar + ar + ar +  diverges if (b) r ≥ 1

(a) |r| < 1 12. The series

1

 np

(a) p > 1 13. The series

1 1 n4

•

(c) p = 0

(d) p = 1

is

(a) convergent (c) oscillates finitely • 32 n 14. The series  is n =1 4 n (a) convergent (c) oscillates finitely 15. The series

(d) r = 1

converges if (b) p < 1

Â

(c) r £ –1

1

 sin n

(b) divergent (d) oscillates infinitely

(b) divergent (d) oscillates infinitely

is

n =1

(a) convergent (c) oscillatory 16.

lim

nÆ a

un = 1 and  vn diverges then  un is vn

(a) divergent (c) oscillatory 2

(b) divergent (d) may or may not be convergent

3

p

p

+

4

+  • converges if 1 2 3p (a) p < 2 (b) p > 2 (c) p > 1 (d) p ≥ 1 18. If Sun is a series of positive terms such that lim un = 0 then Sun is 17.

+

(b) convergent (d) oscillates infinitely

n Æ•

(a) convergent (c) may or may not be convergent

(b) divergent (d) oscillatory

Multiple Choice Questions

19.

1

 n(log n) p

1.113

is convergent

(a) for p > 1 (c) for all real values of p

(b) for p < 1 (d) for no value of p

•

20.

 (2 x)n

is divergent if

n=0

(a) -1 £ x £ 1 (b) -

1 1

•

26. The series

1 3

(c) -

(a) oscillatory (b) divergent •

(d)

1 < x 0, r > 0 (S) Failure Case (iv) rt – s2 > 0, r < 0 22. The function z = 5 xy - 4 x 2 + y 2 - 2 x - y + 5 has at x =

(a) P – i, Q – iii, R – iv, S – ii (c) P–iv, Q – iii, R–ii, S – i

(b) P – ii Q – i, R – iii, S – iv (d) P – iv, Q – ii, R – i, S – iii

8.120

Chapter 8 Applications of Partial Derivatives

24. The minimum value of f(x,y) = x2y2 is [Winter 2015] (a) 1 (b) 2 (c) 4 (d) no conclusion 25. The sum of the squares of two positive numbers is 200, their minimum product is (a) 200 (b) 25 7 (c) 28 (d) 0 26. The coefficient of (x + y)3 in the expansion of sin (x + y) is 1 1 1 1 (a) (b) (c) (d) 6 3 3 6 y 27. The coefficient of x – 1 in the expansion of x near the point (1, 1) is (a) 1 (b) –1 (c) 0 (d 2 28. The coefficients of x – 1 and y + 1 in the expansion of ex + y upto first degree terms are (a) (1, 1) (b) (1, –1) (c) (–1, 1) (d) (0, 1) 29. The coefficient of (x – 1)2 in the expansion of sin xy in powers of (x – 1) and pˆ Ê ÁË y - 2 ˜¯ is p -p p2 -p 2 (b) (c) (d) 8 8 8 8 30. The minimum value of x2 + y2 + z2 given that xy + yz + zx = 3a2 is 1 2 a (a) 3a (b) 4a2 (c) (d) 3a2 3 (a)

31. For the auxiliary equation F ( x, y, z ) = f ( x, y, z ) + lf ( x, y, z ) = 0, the Lagrange’s equations are [Winter 2014] (a)

∂F ∂F ∂F = 0, = 0, =0 ∂x ∂y ∂z

(b)

∂f ∂f ∂f = 0, = 0, =0 ∂x ∂y ∂z

(c)

∂f ∂f ∂f = 0, = 0, =0 ∂x ∂y ∂z

(d)

∂f ∂f ∂f = 0, = 0, =0 ∂x ∂f ∂z

32. The equation of the tangent plane of z = x at (2, 0, 2) is (a) z = x (b) x + y + z = 2 (c) z + x = 0 (d) x + y = 2 33. A point (a, b) is said to be saddle point if at (a, b) [Summer 2016] (a) rt – s2 > 0 (b) rt – s2 = 0 (c) rt – s2 < 0 (d) rt – s2 ≥ 0 34. The minimum value of f(x, y) = x2 + y2 is [Winter 2016] (a) 1 (b) 2 (c) 4 (d) 0 35. If x = u + 3v, (a) –1

∂( x, y) is ∂(u, v) (c) 5

y = –u + v then J = (b) 4

[Winter 2016; Summer 2017] (d) 7

Answers

8.121

Answers 1.(b) 2.(b) 3.(c) 4.(a) 5.(a) 6.(b) 7.(a) 8.(b) 9.(b) 10.(a) 11.(a) 12.(a) 13.(c) 14.(b) 15.(a) 16.(a) 17.(b) 18.(c) 19.(b) 20.(b) 21.(b) 22.(b) 23.(c) 24.(d) 25.(d) 26.(c) 27.(a) 28.(a) 29.(d) 30.(d) 31.(a) 32.(a) 33.(b) 34.(d) 35.(b)

Unit 6 Multiple Integrals

9

CHAPTER

Multiple Integrals 9.1

INTRODUCTION

Integration of functions of two or more variables is normally called multiple integration. The particular case of integration of functions of two variables is called double integration and that of three variables is called triple integration. Sometimes, we have to change the variables to simplify the integrand while evaluating the multiple integrals. Variables can be changed by substitution or by changing the coordinate system (polar, spherical or cylindrical coordinates). Multiple integrals are useful in evaluating plane area, mass of a lamina, mass and volume of solid regions, etc.

9.2

DOUBLE INTEGRALS OVER RECTANGLES

Let f ((x, y) be a continuous function defined in a closed and bounded region R in the xy-plane. Divide the region R into small elementary rectangles by drawing lines parallel to coordinate axes. Let the total number of complete rectangles which lie inside the region R be n. Let dA d r be the area of rth rectangle and (xr, yr) be any point in this rectangle. n

S = ∑ f ( xr , yr )δ Ar

Consider the sum

…(1)

r =1

where d Ar = d xr◊d yr If we increase the number of elementary rectangles then the area of each rectangle decreases. Hence, as n Æ •, d Ar Æ 0. The limit of the sum given by the Eq. (1), if it exists, is called the double integral of f (x, y) over the region R and is denoted by

y dA A

)d A. ∫∫ f ( x, yy))d R

n

Hence,

∫∫ f ( x, yy)))dd A = lim ∑ f ( x , y )δ A R

where d A = d x dy dy

n→∞

δ Ar →0

r

r =1

r

r

x

O

Fig. 9.1

9.2

Chapter 9

9.2.1

Multiple Integrals

Evaluation of Double Integrals by Fubini's Theorem

Double integral of a function f (x, y) over a region R can be evaluated by two successive integrations. There are two different methods to evaluate a double integral which are known as Fubini's theorem. Method-I Let the region R, i.e., PQRS be bounded by the curves y = y1 (x), y = y2 (x) and the lines x = a, x = b. In the region PQRS, draw a vertical strip AB. Along the strip AB, y varies from y1 to y2 and x is fixed. Therefore, the double integral is integrated first w.r.t. y between the limits y1 and y2 treating x as constant. Now, move the strip AB horizontally from PS (i.e., x = a) to QR (i.e., x = b) to cover the entire region PQRS. The result of the first integral is integrated w.r.t. x between the limits a and b. Hence,

∫∫ f ( x, yy))d x ddyy = ∫ ∫ R

b

y2 ( x )

a

y1 ( x )

f ( x, yy))dy  d x 

y

S

R

d

x2 ( y )

c

x1 ( y )

R

x=b

P A

y = y1 (x))

Q x

O

Fig. 9.2 y

S

A In the region PQRS, draw a horizontal strip x =x1(y ) AB. Along the strip AB, x varies from x1 to x2 P and y is fixed. Therefore, the double integral is integrated first w.r.t. x between the limits x1 O and x2 treating y as constant. Now, move the strip AB vertically from PQ (i.e., y = c) to RS (i.e., y = d ) to cover the entire region PQRS. The result of the first integral is integrated w.r.t. y between the limits c and d.

∫∫ f ( x, yy))d x ddyy = ∫ ∫

y = y2 (x))

x=a

Method-II Let the region R, i.e., PQRS be bounded by the curves x = x1( y), x = x2( y) and the lines y = c, y = d.

Hence,

B

y=d

R B x = x2 (y) y=c

Q x

Fig. 9.3

f ( x, yy))d x ddyy 

Note: (i) If all the four limits are constant then the function f (x, y) can be integrated w.r.t. any variable first. But if f (x, y) is implicit and is discontinuous within or on the boundary of the region of integration then the change of the order of integration will affect the result. (ii) If all the four limits are constant and f (x, y) is explicit then double integral can be written as product of two single integrals.

9.2

Double Integrals Over Rectangles

9.3

(iii) If inner limits depends on x then the function f (x, y) is integrated first w.r.t. y and vice-versa.

9.2.2

Properties of Double Integrals

Various properties of double integrals are analogous to those for single integrals. For f and g continuous in region R with k as rational number, (i) (ii)

∫∫ ( f + g ) dxdy = ∫∫ f dxdy + ∫∫ g dxdy ∫∫ k f dxdy = k ∫∫ f dxdy , where k is a constant. R

R

R

R

R

For f continuous in region R, where R = R1 » R2 where R1 and R2 are non-overlapping regions whose union is R: (iii)

∫∫

R

f dxdy = ∫∫ f dxdy + ∫∫ f dxdy R1

R2

Example 1 Evaluate

3

1

0

0

∫ ∫ (x

2

+ 3 y 2 )dy d x.

Solution 3

1

0

0

∫ ∫ ((xx

2

3

+ 3 y 2 ) dy d x = ∫ x 2 y + y 3 0 d x 0 1

3

= ∫ ( x 2 + 1)d x 0

3

x3 +x 3 0 = 12 =

Example 2 1

Evaluate

∫∫

2

0 0

( x 2 + y 2 )dy d x.

Solution

2

1

∫∫

2

0 0

1

((xx 2 + y 2 ) dy d x = ∫ x 2 y + 0

y3 dx 3 0

1 8 = ∫  2x2 +  dx 0 3 1

2 x3 8 x = + 3 3 0 =

10 3

9.4

Chapter 9

Multiple Integrals

Example 3 Evaluate

1

2

Ú-1 Ú0 (1 - 6 x

2

y) dx dy .

[Winter 2016]

Solution 1

2

Ú-1 Ú0 (1 - 6 x

2

1 2 y) dx dy = Ú ÈÍ Ú (1 - 6 x 2 y) dx ˘˙ dy -1 Î 0 ˚

=Ú

1 -1

=Ú

1

-1

x3 x - 6y ◊ 3 2

x - 2 yx 3

0

2

dy 0

dy

1

= Ú (2 - 16 y) dy -1 1

= 2 Ú 2 dy 0

=4

Example 4 Evaluate

a

b

2

2

∫∫

d x dy . xy

Solution a

b

Ú2 Ú2

aÊ b d a dxx dy d dxx ˆ d dyy b 1 = Ú ÁÚ = Ú log x 2 dy ˜ 2 Ë 2 x ¯ y 2 xy y

= (log b - logg 2)Ú

a

2

1 dy y

Ê bˆ a = log g Á ˜ log y 2 Ë 2¯ Ê bˆ = log g Á ˜ (logg a - log 2) Ë 2¯ Ê bˆ Ê aˆ = log g Á ˜ log Á ˜ Ë 2¯ Ë 2¯

Another method: Since both the limits are constant and integrand (function) is explicit in x and y, the integral can be written as a

b

Ú2 Ú2

a dy b d x d x dy =Ú 2 y Ú2 x xy a

b

= log g y 2 log x 2 = (log a - log 2)(log b - logg 2)

9.2

Double Integrals Over Rectangles

9.5

 a  b = logg   ⋅ log    2  2  b  a = logg   ⋅ log    2  2

Example 5 1

2

Evaluate ∫ ∫ xy d y d x. 0 1 Solution 1

∫∫

2

0 1

xy dy dx = ∫

1

=∫

1

0

{ ∫ y dy } x d x 2

1

2

y2 x dx 2 1

0

1 4 1 = ∫  −  x dx 02 2

3 x2 2 2

=

1

0

3 1 ⋅ 2 2 3 = 4 Another method: Since both the limits are constant and integrand (function) is explicit in x and y, the integral can be written as =

1

∫∫

2

0 1

1

2

xy dy d x = ∫ x d x ⋅ ∫ y dy 0

1

2 1

=

x 2

0

2 2

y 2

1

1  4 1  −  2  2 2 3 = 4 =

Example 6 Evaluate

1

∫∫

x

0 0

dy d x.

Solution 1

∫∫

x

0 0

1

x

0

0

dy d x = ∫ y d x

9.6

Chapter 9

Multiple Integrals

1

= ∫ x dx 0

=

1

x2 2

0

1 = 2

Example 7 Evaluate

1

∫∫

x

0 0

y

e x ddyy d x.

Solution 1

∫∫

x

0 0

y

e x dyd x = ∫

1

xe

0

y x x 0

dx

1

= ∫ x(e − 1)dxx 0

1

x2 (e − 1) 2 0 1 = (e − 1) 2 =

Example 8 Evaluate

1

x2

0

x

∫∫

xy dy dx.

Solution 1

x2

0

x

∫∫

xxyy dy dx = ∫

1

=∫

1

0

0

{∫ } x2

y dy x dx

x

y2 2

x2

x dx x

1 1 2 2 ( x ) − x 2  x dx 2 ∫0  1 1 = ∫ ( x 5 − x 3 )dx dx 2 0 =

1 x6 x4 = − 2 6 4 1  1 1 =  −  2  6 4 1 =− 24

1

0

9.2

Double Integrals Over Rectangles

Example 9 1+ x 2

1

Evaluate

∫∫

0 0

dx d y . 1 + x2 + y 2

Solution dy dy  1+ x2  1 y 1 tan −1 dx = ∫ ∫0 ∫0 ( 1 + x 2 )2 + y 2  dx 2 0 1 + x2 1+ x  

1+ x 2

1

=∫

1

=∫

1

0

0

1

(tan −−11 1 − tan −1 0)dx 1 + x2 1 π ⋅ dx 2 1+ x 4

1 π log ( x + 1 + x 2 ) 0 4 π = log (1+ 1+ 2) 4

=

Example 10 1− y 2 2

1

Evaluate

∫∫

dxdy 1 − x2 − y 2

0 0

.

Solution 1

∫∫ 0

0

1− y 2 2

1− y 2 1 = ∫ ∫ 2 0  0 1 − x2 − y 2

d dy dx dy

  dy dy (1 − y 2 ) − x 2  dx dx

1

x

0

1 − y2

−1 = ∫ sin

dx dx 0

1− y 2 2

dy dy 0

1 1  n −−11 − sin sin −1 0 dy = ∫  sin d 0  2 π 1 = y0 4 π = 4

9.7

9.8

Chapter 9

Multiple Integrals

Example 11 1 1 x− y x− y dyy ≠ ∫ dy ∫ d ddxx. 3 0 ( x + y) 0 0 ( x + y )3

1

∫ dx∫

Show that

0

1

Solution 1

1 1 2 x − ( x + y) x− y dyy = ∫ dx ∫ d ddyy 3 0 ( x + y) 0 0 ( x + y )3

∫ dx∫ 0

1

=∫

1

∫

0

1

0

1   2x  ( x + y )3 − ( x + y ) 2  dy dx   1

1 1 1   = ∫ 2x  dx 2+ 0 − 2 ( x + y ) x + y0   1 x 1 1 1 + + −  dx = ∫ − dx 2 0 x + 1 x x ( x + 1 )  1 1 =∫ dx dx 0 ( x + 1) 2 1

1 x +1 0

= − = 1

1

0

0

∫ ddyy ∫

1 2

1 1  ( x + y) − 2 y  x− y dxx = ∫ dy ∫  d dx 3 3  dx 0 0 ( x + y)  ( x + y) 

=∫

1

0

∫

1

0

1

=∫ − 0

2y   1  ( x + y ) 2 − ( x + y )3  dx dy   1

1 y + dy dy x + y ( x + y)2 0

1 1 y 1 1 + + −  dy = ∫ − dy 2 0 y y  1 + y (1 + y ) 1 1 =∫ − dy dy 0 (1 + y ) 2

=

1 1+ y

1 =− 2

1

0

9.2

1 1 x− y x− y dyy ≠ ∫ dy ∫ d ddxx 3 0 ( x + y) 0 0 ( x + y )3

1

∫ dx∫

Hence,

9.9

Double Integrals Over Rectangles

0

1

x− y is discontinuous at (0, 0), a point on the boundary of the region (square), ( x + y )3 change of order of integration does not give the same result.

Since

Example 12 Sketch the region of integration and evaluate

4

Ú1 Ú0

x

3 e 2

y x

dy dx .

[Summer 2017] Solution 1. Since the inner limits depend on x, the function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y: y = 0 to y = x Limits of x: x = 1

to

x=4

3. The region is bounded by the line x = 1, y = 0, x = 4 and parabola y2 = x. 4. The points of intersection of y2 = x and x = 1, x = 4 are (1, 1) (4, 2). 4

Ú1 Ú0

x

3 e 2

y x

4

dy dx = Ú

1

3 ÈÍ x e 2 Í Ú0 Î

y x

4

1

y

y2 = 4 (1, 1)

B

3 e x 2 1/ x

dx

O

A x=1

0

=Ú

4

1

3 x e 2

x

4 x

dx 0

=Ú

4

1

=

(4, 2)

x

y

=Ú

˘ dy ˙ dx ˙ ˚

3 x (e - 1)dx 2

4 3 (e - 1) Ú x dx 1 2

Fig. 9.4

x x=4

9.10

Chapter 9

Multiple Integrals

=

4

3 x2

3 (e - 1) 2

3 2

1 3

3 2 = (e - 1) ◊ x 2 2 3

4

1

3 ˘ = (e - 1) ÈÍ 2 4 Î - 1˙˚

= (e - 1) [2

2◊

3 2

- 1]

3

= [e – 1] [2 – 1] = (e – 1) [8 – 1] = 7(e – 1)

EXERCISE 9.1 Evaluate the following integrals:

∫∫

2. 3.

∫∫

4.

5.

6.

2−y

2

1.

1

− 2−y

∫∫

1

y

1

x

xy e x −2 dx dy

0 0

0 0

1

∫ ∫

2 x 2 y 2 dx dy

e x + y d x dy

1   Ans. : 4e  1  2  Ans. : 2 (e − 1) 

1 y

ye xy dx dy

10 0

[Ans. : 9(1 – e)]

log 8

log y

1

0

∫ ∫ 1

y

0

y2

∫∫

856    Ans. : 945 

e x + y d x dy

(1 + xy 2 )dx dy

[Ans. : 8(log 8 – 1)] 41    Ans. : 210 

9.2

7.

2a

∫ ∫ 0

2 ax − x 2

0

xyy dy dx

Double Integrals Over Rectangles

9.11

 2a 4   Ans. :  3  

9.2.3 Working Rule for Evaluation of Double Integrals Over a General Region 1. If the region is bounded by more than one curve then find the points of intersection of all the curves. 2. Draw all the curves and mark their point of intersection. 3. Identify the region of integration. 4. Draw a vertical or horizontal strip in the region whichever makes the integration easier. 5. The vertical strip starts from the lowest part of the region and terminates on the highest part of the region. 6. For vertical strip: (i) The lower limit of y is obtained from the curve where the vertical strip starts and the upper limit of y is obtained from the curve where it terminates. (ii) The lower limit of x is obtained from the leftmost point of the region and the upper limit of x is obtained from the rightmost point of the region. 7. The horizontal strip starts from the left part of the region and terminates on the right part of the region. 8. For horizontal strip: (i) The lower limit of x is obtained from the curve where the horizontal strip starts and upper limit is obtained from the curve where it terminates. (ii) The lower limit of y is obtained from the lowest point of the region and the upper limit of y is obtained from the highest point of the region. 9. If variation along the strip changes within the region then the region is divided into parts.

Example 1 Evaluate

∫∫ e

ax + by

dx dy, dy over the triangle bounded by x = 0, y = 0, ax + by = 1.

Solution 1. The region of integration is the DOPQ. 2. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the line ax + by = 1. 1 − ax 3. Limits of y : y = 0 to y = b Limits of x : x = 0

to

x=

1 a

9.12

Chapter 9

1

Multiple Integrals

y

1

I

x y

∫ ∫

1 a

1 a

1

e ax

eby d 1 ax

e

ax

eby b

ax + by = 1 B

0

1 a

1 e ax e 1 b∫

1 b

(0, 1 ) b

x

1] dx

ax

( , 0)

1

A

O

dx

x

Fig. 9.5

1

1 ex b

e a a 0 1 e e 1 − + b a a a 1 ab

Example 2 Evaluate

∫∫

Solution

xy 1− y

dx y over the first quadrant of the circle x2 + y2 = 1.

2

1. The region of integration is OPQ. 2. The integration can be done w.r.t any variable first. Draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the circle x2 + y 2 = 1. 3. Limits of y : y = 0 to Limits of x : x = 0 to I

1

1 x2

1 y2 1

1− x 2

1 1 x 2∫ 1 1 − ∫ 2 −

1 − 2

y x=1

x

y

(0, 1) B x2 + y2 = 1

2

dx y (1, 0) −

− 1

1

1

−2 y

x

Fig. 9.6

1 x2

x 1)dx

dx

A

O

∵∫

( x)]n f ( )dx

( x)]n +1  n +1

9.2

x3 3 1 =− 3 1 = 6

9.13

1

x2 2 1 2

=−

Double Integrals Over Rectangles

0

Example 3

∫∫(

Evaluate

) 2 dx y over the right half of the circle x2 + y2 = a2.

Solution 1. The region of integration is PQR. 2. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from the part of the circle a 2 below x-axis and terminates on the part of the y circle a 2 above x-axis. 3. Limits of 2

y y

2

Limits of x : x = 0 a

I

a

2

x

2

=

−x a

x + y2

2

a2 x2 2

−

y−

d

2

Q (a, 0)

2

x

⋅ 2 a2 )

Putting x When x

a co =

When x

x

O

a2 x2

a

x 2 dx

2

A

x 2 dx

P

d

Fig. 9.7

2 d

0

=

+ sin

c

2

n cos 2

π

=2

4

a2

dx y

+

I

B

2

− )2 ∫

0

R

to x = a

a2 x2

=

− x2

s

sin

s

c

2

( sin

d



9.14

Chapter 9

Multiple Integrals

 cos 4  1 π ⋅ + ⋅ ⋅ +2 2 2 4 2 2 3   Using re nf la and

4

16

p p + 2 8

4

5p

a4

-

2

0

[ f ( )]n +1 

d

n +1

˘ p - cos 0 2

3 4 3

4 3

Example 4 n 2

x2 y 2 Evaluate ∫ ∫ xy 2 + 2 a b x2 y 2 + =1 a2 b

over the first quadrant of the ellipse

Solution

y

1. The region of integration is OPQ. 2. The integration can be done w.r.t any variable first. Draw a vertical strip AB parallel to y-axis which starts from x-axis and 2 x2 terminates on the ellipse 2 + =1 a b 3. Limits of y : y = 0

to

Limits of x : x = 0

to

I

a

b 1

x2

0

a

∫ x∫ b2 2

∫

a

b 1

x2 a2

B

x2 a2

x2 a2 2

a2

b2

O

n 2

d

+

y2 b2

=1

P (a, 0)

x=a

2

y

A

x

Fig. 9.8

n

b2 x 2 y 2 + 2 a2 b2 1

x

(0, b)

n +1 2

x2 y2 + a2 b2

2y d b2 1

n +1 2

x2 a2

dx 0

∵ ∫ [ f y )]n f

y) y

[ f y )]n +1  n +1

9.2

9.15

Double Integrals Over Rectangles

b2 x2 1 xn+ 4 − n+ 2 ⋅ n+2 2 a n+4 0 b2 a 2 1 an+ 4 ⋅ n + 2 2 an+ 2 n + 4 a 2 ( + ) ⋅ ( n + 2) 2 + ) 2

a

+ )

2

Example 5

∫∫(

Evaluate

)

over the ellipse 2x2 + y2 = 1.

Solution 2

1. The region of integration is PQRS, the ellipse 2

2

1 or

2

1 2 1 and 1 as its axes. 2 2. The integration can be done w.r.t any variable first. Draw a vertical strip AB parallel to y-axis which starts from the part of the ellipse 2 1 below x-axis and terminates on the part of the ellipse 2 1 above x-axis. 3. Limits of

Limits of x x 1

1 2x

I 2

−

1 2

1 2

2x 2 + y 2

(

y3 3

dx 1 2 x2

− x2 +

2 1

− x2 +

1 3

− 2x

1 − 3

x

)

3 2



dx

1

x

1 0 √2

(√21 , 0)

)

S (0, 1)

A

Fig. 9.9 1− 2 x 2

1

0

to x =

B

dy

1− 2 x 2

+ 2 1

x2

Q (0, 1)

2

1

2

=

2

1

y

with

y y = − 1−

1

9.16

Chapter 9

Multiple Integrals

Putting 2 x 2 = cos 2 θ , x =

1 1 cos θ , dx = − cos sin θ dθ 2 2

When x = 0, cos θ = 0, θ =

π 2

1 , cos θ = 1, θ = 0 2

When x =

3˘ 2 0 È cos q Ê 1 ˆ 1 I = 4Ú p Í 1 - coss 2 q + (1 - cos 2 q ) 2 ˙ Á sin q dq ˜ 2 3 ¯ 2 ˙˚ Ë 2Í Î p È1 p ˘ 1 = 2 2 Í Ú 2 cos cos 2 q sin 2 q dq + Ú 2 sin sin 4 q dq ˙ 0 0 3 ÍÎ 2 ˙˚

È1 Ê 1 1 p ˆ 1 Ê 3 1 p ˆ˘ = 2 2 Í Á ◊ ◊ ˜ + Á ◊ ◊ ˜˙ Î2 Ë 4 2 2¯ 3Ë 4 2 2¯˚ =

Using reductio reductionn fo forrmul mula ] [ Using

3 2p 16

Example 6 Evaluate ∫ ∫ ( x 2 − y 2 )dx dy over the triangle with the vertices (0, 1), (1, 1), (1, 2). Solution 1. The region of integration is DPQR. 2. Equation of the line PQ is y = 1. Equation of the line PR is 2 −1 y −1 = ( x − 0) = x 1− 0 y = x +1 3. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from the line y = 1 and terminates on the line y = x + 1. 4. Limits of y : y = 1 to y = x + 1 Limits of x : x = 0 to x = 1

y R (1, 2) B y=

P (0, 1)

x

+1

A y=1

Q (1, 1)

x

O

Fig. 9.10

9.2

I=∫

1

∫

x +1

0 1

9.17

Double Integrals Over Rectangles

((xx 2 − y 2 )ddyy dx

1

= ∫ x2 y − 0

y3 3

x +1

dx dx 1

1 ((x + 1)3 1 = ∫  x2 ( x + d +1 1 ) − − x 2 +  dx 0 3 3 

1

x 4 x 3 ( x +1 + 1) 4 x 3 x + − − + 4 3 12 3 30 1 1 16 1 = + − + 4 3 12 12 2 =− 3 =

Example 7 Evaluate

∫∫e

y2

dx dy dy over the region bounded by the triangle with vertices

(0, 0), (2, 1), (0, 1).

y

Solution 1. The region of integration is DOPQ. 2. Equation of the line OQ is x y = or x = 2 y. 2 3. Here, it is easier to integrate w.r.t. x first than y. Draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line x = 2y. 4. Limits of x : x = 0 to x = 2y Limits of y : y = 0 to y = 1 I=∫

1

0

∫

1

2y

0

= ∫ ey

2

0

2

1

2

Q (2, 1)

A

B

2y x= x

O

Fig. 9.11

2

e y dxdy

∫

2y

0

1

P (0, 1)

dx dy dy

= ∫ e y x 02 y dy dy 0 = ∫ e y ⋅ 2 y dy dy 0

= e y2

1 0

= e −1

∵ e f ( y ) f ′ ( y )dy = e f ( y )   ∫ 

9.18

Chapter 9

Multiple Integrals

Example 8 2 xy 5

∫∫

Evaluate

over the triangle having vertices

1+ − y4 (0, 0), (1, 1) and (0, 1). Solution

1. The region of integration is the DOPQ. 2. Equation of the line OP is y = x. 3. Here, it is easier to integrate w.r.t. x f irst than with y. Draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. 4. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1 I=Ú =

2 xy 5

1 y

Ú

1+ x

- y4

y

(0, 1) (1, 1)

A

B

dx y

x

O 1

1

2 xy 2 dx y

x

1

1 3

+x y

y

y 0

dy

∵ Ú f x )]n f x )dx =

[ f x )]n +1 n +1

∵ Ú [ ( x )]n f ( )d x

[ ( x )]n +1 ˘ n +1

1

◊

) 2 dy 1

y4 4

1 )2

Fig. 9.12

1

-2

0 1

1

1

dy 1

1 - y )2

0

1 1 2 + (1 2 2 3 1 1 2 1 6

3 )2

( 4y ) dy -4

1

0

9.2

9.19

Double Integrals Over Rectangles

Example 9 Evaluate ∫ ∫(( x 2 + y 2 )dx dy over the region bounded by the lines y = 4x, x + y = 3, y = 0, y = 2. Solution y

1. The region of integration is OPQR. 2. The integration can be done w.r.t. any variable first. But in case of vertical strip we need to divide the region into three parts. Therefore, draw a horizontal strip AB parallel to x-axis which starts from the line y = 4x and terminates on the line x + y = 3.

R A

to

B

x=3–y P

O

to

y=2

2 3- y 2 y (x 0 4

I=Ú =Ú

Ú

2

0

+ y 2 )dx dy

x3 + xy 2 3

Fig. 9.13

3- y

dy y 4

3 2 È (3 - y ) 1 y3 y3 ˘ =Ú Í + (3 - y) y 2 - ◊ - ˙ dy 0 3 64 4 ˙˚ ÍÎ 3 3 2 È (3 - y ) 241 3 ˘ =Ú Í + 3 y2 y ˙ dy 0 19 92 2 ˚˙ ÎÍ 3

1 (3 - y)4 y3 241 y 4 = ◊ + 3◊ ◊ 3 -4 3 192 4 ==

y=2

Q

x+y=3

y 3. Limits of x : x = 4 Limits of y : y = 0

y = 4x

1 241 Ê 27 ˆ +8◊4 - Á- ˜ Ë 4¯ 12 192

463 48

2

0

x

9.20

Chapter 9

Multiple Integrals

Example 10 Evaluate

Ú Ú ( x + y)dy dx, where R is the region bounded by x = 0, x = 2, R

y = x, y = x + 2.

[Summer 2016]

Solution 1. The region OPQR.

of

integration

is y

2. The point of intersection of x = 2 and y = x + 2 is obtained as y = 4. The point of intersection is (2, 4). 3. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from the line y = x and terminates on the line y = x + 2. 4. Limits of y: y = x to y = x + 2 Limits of x: x = 0 to x = 2 2

I = ÚÚ 0

x+2 x

Q

y= R x=0

x+

(

y= O

0

Fig. 9.14

x+2

dx x

È 1 x2 ˘ = Ú Í x( x + 2) + ( x + 2)2 - x 2 - ˙ dx 2 2 ˙˚ Í 0Î 2 È x2 x2 ˘ = Ú Í x2 + 2 x + + 2 x + 2 - x 2 - ˙ dx 2 2 ˙˚ Î 0Í 2

2

= Ú [4 x + 2] dx 0

x2 = 4 + 2x 2 = 2 x2 + 2 x = 8+4-0 = 12

2

0 2 0

(2, 2)

x=2 x

( x + y) dy dx

y2 2

x

(–2, 0)

2

2

P

)

2 0,

x+2 = Ú ÈÍ Ú ( x + y) dy ˘˙ dx x Î ˚ 0

= Ú xy +

(2, 4)

2

9.2

Double Integrals Over Rectangles

9.21

Example 11

ÚÚ (2 x - y

2

) dA A over the triangular region R enclosed between the lines

R

y = –x + 1, y = x + 1 and y = 3.

[Summer 2015]

Solution 1. The region of integration is DPQR. 2. The points of intersection of (i) y = –x + 1 and y = x + 1 is obtained as –x + 1 = x + 1 2x = 0, x = 0 y=1 The points of intersection is P(0, 1). (ii) y = x + 1 and y = 3 is obtained as 3=x+1 x = 2, y = 3 The points of intersection Fig. 9.15 is Q(2, 3). (iii) y = –x + 1 and y = 3 is obtained as 3 = –x + 1 x = –2, y = 3 The points of intersection is R(–2, 3). 3. The integration can be done w.r.t. any variable first. But in case of vertical strip, we need to divide the region into two parts. Therefore, draw a horizontal strip AB parallel to x-axis which starts from the line y = –x + 1 and terminates on the line y = x + 1. 4. Limits of x : x = 1 – y to x = y – 1 Limits of y : y = 1 to y = 3

I=Ú

3

=Ú

3

y -1

((2 2x - y 1 Ú1- y

Ú

y -1

1 1- y

2

((2 2 x - y 2 ) dx dy

3

y -1

1

1- y

= Ú x 2 - xxyy 2 3

) dA

dy

= Ú ÈÎ( y - 1)2 - y 2 ( y - 1) - ((1 1 - y)2 + y 2 (1 - y)˘˚ dy 1

9.22

Chapter 9

Multiple Integrals

3

dy

1 3

3

-

3

3

4

3

4

3

3

8 81 54 1 2 ˘ - + + 3 2 3 2 3

8 3 68 3

Example 12 Evaluate

y

∫∫ (

y2

)

over the region bounded by the parabola

y 2 = x and the line y = x. Solution 1. The region of integration is OPQ. 2. The points of intersection of y2 = x and y = x are obtained as x2

y B

y2

x (1, 1)

x Q

A

x 01 \ =0 1 The points of intersection are O (0, 0) and P (1, 1). 3. Here, it is easier to integrate w.r.t. y first than x. Draw a vertical strip AB parallel to y-axis, which starts from the line y = x and terminates on the parabola y2 = x. 4. Limits of y : y = x

to

y

Limits of x : x = 0

to

x=1

I=Ú

1

=Ú

1

Úx

Fig. 9.16

y2

(a x ) 1

x

y

x

(a x ) Ú

x

-

1 2

x

O

1

ax

9.2

=-

1 1 1 1 2 2 2 Ú0 ((aa - x ) 2((ax - y )

= -Ú

1

= -Ú

1

0

0

1 ((aa - x )

x

dx x

9.23

Double Integrals Over Rectangles

È [ f ( y)]n +1 ˘ n ¢( y)dy = Í∵ Ú [ f ( y)] f ¢( ˙ n +1 ˚ Î

1 1˘ È 2 ÎÍ((ax - x ) 2 - (ax - x ) 2 ˚˙ dx

x ( a --11 - a - x ) dx dx a-x

Putting x = a sin2 q, dx = 2a sin q cos q dq When x = 0, q = 0 1

When x = 1, q = sin -1 I = -Ú

a

1

sin -1

a

0

= -2 a Ú

a sin q a cos2 q 1

sin -1

a

0

= -2 a Ú

sin -1

1 a

0

= -2 a Ú

sin -1

0

= -2 a

= -2 a

1 a

(

cos q dq a - 1 - a cos q ) 2 a sinn q cos

sin 2 q ( a - 1 - a cos q ) dq cos q

ÈÊ 1 - cos2 q ˆ 2 ˘ ÍÁ ˜¯ a - 1 - a sin q ˙ dq Ë cos q Î ˚ È ˘ a cos q ) ((11 - cos cos 2q )˙ dq Í a - 1(sec q - cos 2 Î ˚

aq a sin 2q a - 1[log(sec [log(sec q + ttan q ) - sin q ] + 2 4

sin -1

1 a

0

È Ê 1 + sin q ˆ ˘ aq a sin sin q ccos q a - 1 Í log Á + ˜¯ - sin q ˙ Ë 2 2 cos q Î ˚

sin -1

1 a

0

È ˘ 1 Ê ˆ 1 1+ a sin -1 Í ˙ Á ˜ 1 a 1 1˙ a a = -2 a Í a - 1 Á log + ◊ ◊ 1 ˜ Í 2 2 a˙ 1 a˜ a Á 1 Í ˙ ÁË ˜¯ a Î ˚ = -2 a(a - 1) log g

a +1 a -1

+ a - 1 + a sin -1

1 a

9.24

Chapter 9

Multiple Integrals

Example 13 y over the region enclosed by the parabola x2 = y and Evaluate ∫ ∫ the line y = x + 2. y (2, 4) Solution 1. The region of integration is POQ. 2. The points of intersection of x2 = y and y = x + 2 are obtained as x x

2

x+2

2

x2 = y P ( 1, 1)

x 2 0 x 1) = 0 x= 1 ∴ =41

(x

B

A O

x

Fig. 9.17

The points of intersection are P (–1, 1) and Q (2, 4). 3. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from the parabola x2 = y and terminates on the line y = x + 2. 4. Limits of y : y = x2 to y = x + 2 Limits of x : x = –1 to x = 2 I=Ú

2

=Ú

2

=

Ú

x 2

1 x

1

y2 2

x+2

dx x

1 2 2Ú 1 1 2

dx 5 2

3

−

3 5 1 64 32 1 = 2 3 5 3 36 = 5

1

1 5

Example 14 Evaluate ∫ ∫ xy( x x2 = y, y2 = –x.

y)

, over the region enclosed by the parabolas

Solution 1. The region of integration is OPQ.

9.2

Double Integrals Over Rectangles

2. The points of intersection of the parabola x2 = y, and y2 = –x are obtained as y4 = y y = 0, 1 \ x = 0, –1. The points of intersection are O (0, 0) and Q (–1, 1). 3. Here, it is easier to integrate w.r.t. x first. Draw a horizontal strip AB parallel to x-axis, which starts from the parabola x2 = y and terminates on the parabola y2 = –x. 4. Limits of x : x = − y to x = –y 2 Limits of y : y = 0

9.25

y Q (−1, 1) A x2 = y

B y 2 = −x

P O

x

to y = 1 1

I=∫y

∫

0

− y

1

− y2

=∫y 0

1

Fig. 9.18

− y2

∫

xxyy ( x + y )dx dy dy ( x 2 + xy )dx dy

− y

x3 x 2 y =∫y + 3 2 0

− y2

dyy − y

5  7  −y y6 y 2 y3   =∫ + + − dy dy  3 2 3 2 0 1

7

y8 y 7 2 y 2 y 4 = − + + − 24 14 21 8 =0

1

0

Example 15 Evaluate ∫ ∫ xy dx dy over the region enclosed by the x-axis, the line x = 2a and the parabola x2 = 4ay. Solution 1. The region of integration is OPQ. 2. The point of intersection of the parabola x2 = 4ay and the line x = 2a is obtained as 4a2 = 4ay y=a The point of intersection is Q (2a, a). 3. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis, which starts from x-axis and terminates on the parabola x2 = 4ay.

9.26

Chapter 9

Multiple Integrals

4. Limits of y : y = 0

to

Limits of x : x = 0

to

I=∫

2a

0

∫

x2 4a

0

2a

= ∫ x∫ 0

y

x2 4a x = 2a y=

Q (2 (2a, a) x 2 = 4ay

xy d dxx dy

x2 4a

0

B

y dy dx

P (2a, 0)

=∫ x 0

=∫

2a

0

=

x 2 4a

y 2

A

O

2

2a

x = 2a

x

dx dx Fig. 9.19

0

x4 x⋅ dx dx 32a a2

1 x6 32a 2 6

2a

0

1 64a 6 = ⋅ 32a 2 6 a4 = 3

Example 16 Evaluate ∫ ∫ xy dx dy, over the region enclosed by the circle x2 + y2 – 2x = 0, the parabola y2 = 2x and the line y = x. Solution 1. The region of integration is OPQRO. 2. (i) The points of intersection of the circle x2 + y2 – 2x = 0 and the line y = x are obtained as x2 + x2 – 2x = 0 x = 0, 1 \ y = 0, 1 The points of intersection are O (0, 0) and P (1, 1). (ii) The point of intersection of the circle x2 + y2 – 2x = 0 and the parabola y2 = 2x is obtained as x2 + 2x – 2x = 0 x=0 \y=0 The point of intersection is O (0, 0).

D R B

y 2 = 2x Q (2, 2) x = y

C A

P (1, 1)

x 2 + y 2 − 2x = 0

O

Fig. 9.20

9.2

Double Integrals Over Rectangles

9.27

(iii) The points of intersection of the parabola y2 = 2x and the line y = x are obtained as x2 = 2x x = 0, 2 \ y = 0, 2 The points of intersection are O (0, 0) and Q (2, 2). 3. The integration can be done w.r.t. any variable first. To integrate w.r.t. y first we need to draw a vertical strip in the region. But one vertical strip does not cover the entire region, therefore, divide the region OPQRO into two subregions OPR and RPQ and draw one vertical strip in each subregion. 4. In the subregion OPR, strip starts from the circle x2 + y2 – 2x = 0 and terminates on the parabola y2 = 2x. Limits of y : y = 2 x − x 2 to y = 2 x Limits of x : x = 0 to x = 1. 5. In the subregion RPQ, strip starts from the line y = x and terminates on the parabola y2 = 2x. Limits of y : y = x to y = 2 x Limits of x : x = 1 to x = 2 I = Ú Ú xy ddxx dy

ÚÚ

=

xy dx dy +

OPR

=Ú

1

0 1

Ú

2x 2 x - x2

= Ú xÚ 0

xy dx dy

xxyy dy dx + Ú

2

1

2x

Úx

2x

2

2x-x

2 1 =Ú x y 0 2

=

ÚÚ

RPQ

2

y dy dx + Ú x Ú 1

xxyy dy dx

2x x

2x

2 2 dx + Ú x y 1 2 2 x - x2

y dy dx

2x

dx x

1 1 1 2 x(2 x - 2 x + x 2 )dx + Ú x(2 x - x 2 )dx Ú 2 0 2 1 1

1 4 1 2 x3 x 4 = x + 2 4 0 2 3 4 1 8 1 1 = + -2- + 8 3 3 8 7 = 12

2

1

Example 17 dy, over the region in the first quadrant enclosed by Evaluate ∫ ∫ x dxdy the rectangular hyperbola xy = 16, the lines y = x, y = 0 and x = 8. [Winter 2014] 2

9.28

Chapter 9

Multiple Integrals

Solution y 1. The region of integration is OPQR. 2. (i) The points of intersection of the hyperbola xy = 16 and the line y = x are obtained as x = y x2 = 16, x = ± 4 B \y=±4 x=8 Hence, R (4, 4) is the point of interR (4, 4) xy = 16 section in the first quadrant. B D Q (8, 2) (ii) The point of intersection of the hyperbola xy = 16 and line x = 8 is obtained as O A M C P x y=0 8y = 16 y=2 Fig. 9.21 The point of intersection is Q (8, 2). 3. The integration can be done w.r.t. any variable first. To integrate w.r.t. y first we need to draw a vertical strip in the region. But here one vertical strip cannot cover the entire region, and therefore divide the region OPQR into two subregions OMR and RMPQ and draw one vertical strip in each subregion. 4. In the subregion OMR, strip starts from x axis and terminates on the line y = x. Limits of y : y = 0 to y = x Limits of x : x = 0 to x = 4 5. In subregion RMPQ, strip starts from x axis and terminates on the rectangular hyperbola xy = 16 16 Limits of y : y = 0 to y = x Limits of x : x = 4 to x = 8

I = Ú Ú x 2 dx dy dy =

ÚÚ

OMR OM

=Ú

4

0

ÚÚ

x 2 dx ddyy +

x 2 dx dy

RMPQ x

Ú0

x 2 dy ddxx + Ú

4

x

0

0

4 2 x 0 4 3

y 0 dx + Ú

16 x 4 0 8

Ú

x 2 dy dx

16 x dy dx 4 0 16 8 2 x y 0x dx 4 8

= Ú x 2 Ú dy dx + Ú x 2 Ú =Ú

x

8

= Ú x dx + Ú x 2 ◊ 0

4

4

8

4 2 = x + 16 x 4 0 2 4 = 64 + 8(64 64 - 16 16) = 448

16 dx x

9.2

9.29

Double Integrals Over Rectangles

Example 18 Evaluate

dx d y , over the region bounded by the y ≥ x2, x ≥ 1. 4 + y2

∫∫ x

Solution 1. The region of integration is bounded by y ≥ x2 (the region inside the parabola x2 = y) and x ≥ 1 (the region on the right of line x = 1). 2. The point of intersection of x2 = y and x = 1 is obtained as 1 = y. The point of intersection is P(1, 1). 3. Here, it is easier to integrate w.r.t. y first than x. Draw a vertical strip AB parallel to y-axis in the region which starts from the parabola x2 = y and extends up to infinity. 4. Limits of y : y = x2 to y Æ • Limits of x : x = 1 to x Æ • • • 1 I=Ú Ú 2 4 dyydx d 1 x x + y2 •

1

1

x2 1

=Ú

•

tan -1

y x2

y B

A

x2 = y

P (1, 1) x

O

x=1

Fig. 9.22

•

dx dx x2

(tan -1 • - tan -1 1)dx x2 • 1 Êp pˆ = Ú 2 Á - ˜ dx 1 x Ë 2 4¯

=Ú

1

•

p 1 4 x1 p = 4 =

EXERCISE 9.2 Evaluate the following integrals: 1. 2.

1

∫ ∫ xy dx dy, over the rectangle 1 £ x £ 2, 1 £ y £ 2.

[Ans. : (log 2)2]

( + by))dx dy , over the triangle bounded by the lines x = 0, y = 0 ∫∫ sinn π(ax and ax + by = 1.

1    Ans. : π ab 

9.30

3.

Chapter 9

∫∫ e

3x +4 y

Multiple Integrals

dx dy , over the triangle bounded by the lines x = 0, y = 0, and

x + y = 1. 1   4 3  Ans. : 12 (3e − 4e + 1) 4.

∫∫ xy

1− − x − y dx dy, y over the triangle bounded by x = 0, y = 0 and x + y = 1. 16    Ans. : 945 

5.

∫∫

6.

∫ ∫ (x + y + a)dx dy, over the region bounded by the circle x

xy − y 2 dx dy , over the triangle having vertices (0, 0), (10, 1), (1, 1). [Ans. : 6] 2

+ y 2 = a 2. [Ans. : p a3]

7.

∫ ∫ xydxdy, over the region bounded by the x-axis, the line y = 2x and the parabola y =

8.

x2 . 4a

2048 4    Ans. : 3 a 

∫ ∫ 5 − 2x − y (dx) dyy, over the region bounded by x-axis, the line x + 2y = 3

and the parabola y2 = x.

217    Ans. : 60 

1

9.

10.

11.

2 2 ∫ ∫ (4 x − y )2 dx dy, over the triangle bounded by x-axis, the line y = x and x = 1.  1π 3  Ans. :  +  3 3 2  

∫ ∫ xy(x + y)dx dy, over the region bounded by the parabola y

2

∫ ∫ xy(x + y)dx dy, over the region bounded by the curve x

2

line x = y.

12.

= x, x2 = y. 3   Ans. : 28 

= y and the 3   Ans. : 56 

∫ ∫ xy(x − 1)dx dy, over the region bounded by the rectangular hyperbola xy = 4, the lines y = 0, x = 1, x = 4 and x-axis. [Ans. : 8(3 – log 4)]

9.3

9.3

9.31

Change of Order of Integration

CHANGE OF ORDER OF INTEGRATION

Sometimes, evaluation of double integral becomes easier by changing the order of integration. To change the order of integration, first, we draw the region of integration with the help of the given limits. Then we draw a vertical or horizontal strip as per the required order of integration. This change of order also changes the limits of integration.

Type I Change of Order of Integration

Example 1 Change the order of integration of

1

∫∫

x

f x, y x y.

Solution

y

1. Since inner limits depends on x, the function is integrated first w.r.t. y and then w.r.t. x.

Q (1, 1) B′

The correct form of the integral 1

∫∫

x

x x=1

2. Limits of y : y = 0 to y = x, along vertical strip A¢B¢ Limits of x : x = 0 to x = 1 3. The region is bounded by the lines y = 0, y = x, and x = 1. 4. The point of intersection of y = x and x = 1 is Q(1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from the line y = x and terminates on the line x = 1. Limits of x : x = y to x = 1 Limits of y : y = 0 to y = 1 Hence, the given integral after change of order is

∫∫

f x, y

1 y

O

y

0

A′

Fig. 9.23 y Q (1, 1) x

A

P

O

Fig. 9.24

Example 2 1

∫∫

y

1

B

f ( x y )dx y

Change the order of integration of

x

P

x

x

9.32

Chapter 9

Multiple Integrals

Solution

y

1. Since inner limits depend on y, the function is integrated first w.r.t. x and then w.r.t. y. The correct form of the integral

P

=∫

1

∫

y

y

Q (1, 1)

f x, y x y. A′

2. Limits of x : x = 0 to x = y, along horizontal strip A¢B¢ Limits of y : y = 0 to y = 1 3. The region is bounded by the lines x = 0, x = y, and y = 1. 4. The point of intersection of x = y and y = 1 is Q (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from the line x = y and terminates on the line y = 1. Limits of y : y = x to y = 1 Limits of x : x = 0 to x = 1

f x, y x y

1

f ( x y)

B

x

O

Fig. 9.25 y

P

Hence, the given integral after change of order is

∫∫

1

y

B

1

Q (1, 1)

A

x

x

O

Fig. 9.26

Example 3 Change the order of integration of

a

∫∫

a

x

x

Solution y

1. Since inner limits depend on x, the function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y : y = x to y = a, along vertical strip A¢B¢. Limits of x : x = 0 to x = a 3. The region is bounded by the lines y = x, y = a and x = 0. 4. The point of intersection of y = x and y = a is Q (a, a). 5. To change the order of integration i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from

P

B′

y

a

Q (a a)

A′ x

O

Fig. 9.27

9.3

Change of Order of Integration

the line x = 0 and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = a Hence, the given integral after change of order is

∫

a

a

y y=a

P

Q (a a)

A

f x y

9.33

B

f ( x y )dx y x

O

Fig. 9.28

Example 4 ∞

Change the order of integration of

∫ ∫

∞ x

f x, y x y. y

Solution

yƕ

B

1. Since inner limits depend on x, the function is integrated first w.r.t. y and then w.r.t. x. The correct form of the integral =

∞

∫ ∫

∞ x

f x, y

dx

2. Limits of y : y = x to y Æ •, along vertical strip Limits of x : x = 0 to x Æ • 3. The region is bounded by the lines y = x and x = 0. 4. Here the only point of intersection is origin O.

A x

O

Fig. 9.29 y

5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from the line x = 0 and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y Æ • Hence, the given integral after change of order is

∫

∞

∞

f x y

x

∞

y

A

B

x

O

f ( x y )dx y Fig. 9.30

9.34

Chapter 9

Multiple Integrals

Example 5 1

Change the order of integration of

∫∫

x x

f x, y

x.

Solution y

1. The function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y : y = x to y x Limits of x : x = 0 to x = 1 3. The region is bounded by the line y = x and the parabola y2 = x. 4. The points of intersection of y2 = x and y = x are obtained as x2 = x x = 0, 1 \ y = 0, 1. The points of intersection are O (0, 0) and Q (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from the parabola y2 = x and terminates on the line y = x. Limits of x : x = y2 to x = y Limits of y : y = 0 to y = 1 Hence, the given integral after change of order is

∫

x

(1, 1) B′

A′ x

O

Fig. 9.31 y

y A

(1, 1) B

x

O

y y

Fig. 9.32

Example 6 1

1

Change the order of integration of

∫∫

y y2

f x, y x y.

Solution 1. The function is integrated first w.r.t. x and then w.r.t. y.

9.3

Change of Order of Integration

y

1

2. Limits of x : x = y2 to

x

y3

Limits of y : y = 0 to y = 1 3. The region is bounded by the parabola y2 = x and the cubical parabola y = x3. 4. The points of intersection of y2 = x and y = x3 are obtained as x6 = x x = 0, 1 \ y = 0, 1. The points of intersection are O (0, 0) and Q (1, 1).

(1, 1) y2

y

x

x

x3

y

A′ B′

x

O

Fig. 9.33

5. To change the order of integration , i.e., to integrate first w.r.t. y, draw a vertical strip parallel to y-axis which starts from the cubical parabola y = x3 and terminates on the parabola y2 = x. Limits of y : y = x3 to

9.35

y

(1, 1) B y2 = x y = x3

A

Limits of x : x = 0 to x = 1 Hence, the given integral after change of order is

x

O

1

∫

y

x

2

Fig. 9.34

Example 7

y

Change the order of integration of 8

∫∫

y 4 y 8 4

(2, 8)

A′

Solution 1. The function is integrated first w.r.t. x and then w.r.t. y. y 8 y 2. Limits of x : x to x 4 4 Limits of y : y = 0 to y = 8 3. The region is bounded by the line y = 4x + 8, y = 4x, y = 8 and x-axis (y = 0).

y=8

Q (0, 8)

f x, y x y.

R (−2, 0)

B′

O

Fig. 9.35

x

9.36

Chapter 9

Multiple Integrals

4. The point of intersection of y = 4x and y = 8 is obtained as 8 = 4x x = 2. The point of intersection is P (2, 8). 5. To change the order of integration, i.e., to integrate first w.r.t. y, divide the region OPQR into two subregions OQR and OPQ. Draw a vertical strip parallel to y-axis in each subregion.

y

C B

(i) In subregion OQR, strip AB starts R from x-axis and terminates on the line ( 2, 0) A y = 4x + 8. Limits of y : y = 0 to y = 4x + 8 Limits of x : x = –2 to x = 0 (ii)In subregion OPQ, strip CD starts from the line y = 4x and terminates on the line y = 8. Limits of y : y = 4x to y = 8 Limits of x : x = 0 to x = 2 Hence, the given integral after change of order is

∫∫

y y− 4

f x, y x y

+

x

∫

x

Example 8 Change the order of integration of

a

∫ ∫ a

y=8 D

Q (0, 8)

y2 a

8 x

f x, y

x

O

Fig. 9.36

dx

f x, y x y.

Solution 1. The function is integrated first w.r.t. x and then w.r.t. y. y2 2. Limits of x : x = 0 to x = a Limits of y : y = –a to y = a 3. The region is bounded by the y-axis, the parabola y2 = ax, and the line y = – a, and y = a. 4. (i) The point of intersection of y2 = ax and y = –a is obtained as a2 = ax x=a The point of intersection is R (a, –a).

y P A

y=a B′

Q (a a)

y 2 = ax x

O

S

R (a y= a

Fig. 9.37

a)

9.3

9.37

Change of Order of Integration

(ii) The point of intersection of y2 = ax and y = a is obtained as a2 = ax x=a The point of intersection is Q (a, a).

y D

P

y=a Q (a a) y 2 = ax

C

5. To change the order of integration, i.e., to integrate first w.r.t. y, divide the region into two subregions ORS and OPQ. Draw a vertical strip parallel to y-axis in each subregion.

x

O B R (a

S

a)

A y = −a (i) In subregion ORS, strip AB starts from the line y = –a and terminates on the Fig. 9.38 parabola y2 = ax. Limits of y : y = –a to y ax (part of the parabola below x-axis) Limits of x : x = 0 to x = a (ii) In subregion OPQ, strip CD starts from the parabola y2 = ax and terminates on the line y = a. Limits of y : y ax to y = a Limits of x : x = 0 to x = a

Hence, the given integral after change of order is

∫ ∫

y a

f x, y x y

a

ax

x

Example 9 Change the order of integration of

x

2

∫∫

2 y

a

∫

a

4− 2 y

f x, y

dx

f x, y x y.

Solution 1. The function is integrated first w.r.t. x and then w.r.t. y. 2. Limits of x : x = y to x y Limits of y : y = 0 to y = 2 y 3. The region is bounded by the x-axis, the line y = x and the parabola (x – 2)2 = 2(2 – y). 4. The points of intersection of y = x and (x – 2)2 = 2(2 – y) are obtained A′ as O (x – 2)2 = 2(2 – x) x = 0, 2 \ y = 0, 2.

Q (2, 2) (x

2) = 2(2 − y) B′ R (4, 0) x

Fig. 9.39

9.38

Chapter 9

Multiple Integrals

The points of intersection are O (0, 0) and Q (2, 2). 5. To change the order of integration, i.e., to integrate first w.r.t. y, divide the region into two subregions OPQ and PQR. Draw a vertical strip parallel to y-axis in each subregion. y

(i) In subregion OPQ, strip AB starts from x-axis and terminates on the line y = x. Limits of y : y = 0 to y = x Limits of x : x = 0 to x = 2 (ii) In subregion PQR, strip CD starts from x-axis and terminates on the parabola (x – 2)2 = 2(2 – y). Limits of y : y = 0

to

B

x

2

2

(4, 0) O

A

(2, 0) C

x

Fig. 9.40

x2 − 2

y

Q (2, 2) D

Limits of x : x = 2 to x = 4 Hence, the given integral after change of order is

∫∫

− y

x

f x, y x y

∫

x

2 x−

x2 2

f x, y

dx

Example 10

∫

Change the order of integration of

a co

∫

a2 x2

x

tan

Solution 1. The function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y : y = x tan a

to

2

x2

Limits of x : x = 0 to x = a cos a 3. The region is bounded by the line y = x tan a, the circle x2 + y2 = a2 and y-axis. Since given limits of x and y are positive, the region lies in the first quadrant. 4. The points of intersection of y = x tan a and x2 + y2 = a2 are obtained as x2 + x2 tan2a = a2 x = ± a cos a P \ y = ± a sin a . ( a cos a The points of intersection are P − sin a) (a cos a, a sin a ) and P ¢(– a cos a, – a sin a ).

y (0, a) B′

y = x tan a P (a cos a, a sin a )

A′ O

Fig. 9.41

x2 + y 2 x

a2

9.3

Change of Order of Integration

5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into two subregions OPR and PQR. Draw a horizontal strip in each subregion.

y (0, a) D

C R

y = x tan a P (a cos a,

A

(i) In subregion OPR, strip AB starts from y-axis and terminates on the line y = x tan a. Limits of x : x = 0 to x = y cot a P′ Limits of y : y = 0 to (− cos a a sin a) y = a sin a (ii) In subregion PQR, strip CD starts from y-axis and terminates on the circle x2 + y2 = a2. 2 Limits of x : x = 0 to y2 Limits of y : y = a sin a to y = a Hence, given integral after change of order is a2 x 2

9.39

x2+y2

B

sin a ) a2

x

O

Fig. 9.42

ot

a

Ú sin a Ú

a2 y2

y

Example 11 Change the order of integration of

4

∫∫

4x 4 x x2

f x, y

x.

Solution 1. The function is integrated first w.r.t. y and then w.r.t. x. y 2. Limits of y : = x (4, 4) B′ Limits of x : x = 0 to x = 4. 3. The region is bounded by the circle y 2 = 4x x2 + y2 – 4x = 0, the parabola y2 = 4x A and the line x = 4. 4. (i) The point of intersection of (4, 0) x2 + y2 – 4x = 0 and y2 = 4x is O x obtained as 2 x =0 x 2 + y 2 − 4x 0 x=0 \ y = 0. The point of intersection is O (0, 0). (ii) The points of intersection of y2 = 4x Q ′(4, −4) and x = 4 are obtained as y2 = 16 Fig. 9.43 y=±4 The points of intersection are Q (4, 4) and Q ¢ (4, – 4).

9.40

Chapter 9

Multiple Integrals

5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into three subregions ORT, TPS and RSQ. Draw a horizontal strip parallel to x-axis in each subregion. (i) In subregion ORT, strip AB starts from the parabola y2 = 4x and terminates on the circle x2 + y2 – 4x = 0. Limits of x :

y Q (4, 4) F

E

y 2 = 4x

T (2, 2)

R A B

S (4, 2) D

C

P (4, 0) O

(2, 0)

x x 2 + y 2 − 4x = 0

y2 x= to x = 2 − 4 − y 2 4 (Part of the circle where x < 2) Limits of y : y = 0 to y = 2 (ii) In subregion TPS, strip CD starts from the circle x2 + y2 – 4x = 0 and terminates on the line x = 4.

Q' (4, −4)

Fig. 9.44

Limits of x : x = 2 + 4 − y 2 (Part of circle where x > 2) to x = 4 Limits of y : y = 0 to y = 2 (iii) In subregion RSQ, strip EF starts from the parabola y2 = 4x and terminates on the line x = 4. y2 to x = 4 Limits of x : x = 4 Limits of y : y = 2 to y = 4 Hence, given integral after change of order is 4

∫∫ 0

4x 4 x− x x

2

f ( x, y ))d dy dx dx = ∫

2

0

∫

+∫

2− 2 − 4− y2 y2 4 2

0

∫

f ( x , y ) dx d y

4

2+ 2 + 4− y2

f ( x, yy))dx dy dy + ∫

4

2

∫

4 y2 4

f ( x, y )dx dy dy

Example 12 Change the order of integration of

2

((4 − x )2

0

4− x

∫∫

f ( x, y ))ddy dx dx.

Solution 1. The function is integrated first w.r.t. y and then w.r.t. x. 2. Limits of y : y = 4 − x to y = (4 − x) 2 . Limits of x : x = 0 to x = 2 3. The region is enclosed by the parabolas y2 = 4 – x, y = (4 – x)2, the lines x = 0 and x = 2. 4. (i) The points of intersection of x = 2 and y2 = 4 – x are obtained as

9.3

y2 = (4 – 2)

y

y = ± 2. The points of intersection are Q Q (2 ). (ii) The point of intersection of x = 2 and y = (4 – x)2 is obtained as y = (4 – 2)2 = 4. The point of intersection is S (2, 4). (iii) The points of intersection of x = 0 and y2 = 4 – x are obtained as y2 = 4 y = ±2. The points of intersection are P (0, 2) and P¢ (0, –2). (iv) The point of intersection of x = 0 and y = (4 – x)2 is obtained as y = 16. The point of intersection is U (0, 16). 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into three subregions PQR, PRST and STU. Draw a horizontal strip in each subregion. (i) In subregion PQR, strip AB starts from the parabola y2 = 4 – x and terminates on the line x = 2. Limits of x : x = 4 – y2 to x = 2 Limits of y : y= 2 to y = 2 (ii) In subregion PRST, strip CD starts from y-axis and terminates on the line x = 2. Limits of x : x = 0 to x = 2 Limits of y : y = 2 to y = 4

9.41

Change of Order of Integration

U (0, 16) = (4 − x)2 B x=2

P (0, 2)

A′ Q (2, 2)

O

V (4, 0) y2 Q (2,

x

−x

2)

P′ (0, 2)

Fig. 9.45 y U (0, 16)

= (4 − x

F

E T C P (0, 2)

2

D A

R (2, 2) B

Q (2, 2)

x=2 O

V (4, 0) y 2 = (4 − x Q ′(2, − 2) P (0, −2)

Fig. 9.46

x

9.42

Chapter 9

Multiple Integrals

(iii) In subregion STU, strip EF starts from y-axis and terminates on the parabola y = (4 – x)2. Limits of x : x = 0 to x y (Part of the parabola where x < 4) Limits of y : y = 4 to y = 16 Hence, the given integral after change of order is

∫∫ 0

(

x −

f x, y

x

−

+∫

16 4

∫

4

y

∫

dx

2

f x, y x y

f x, y x y

Type II Evaluation of Double Integrals by Changing the Order of Integration

Example 1 Change the order of integration and evaluate

a

∫∫

a x

(

)

.

Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y. 2. Limits of y : y = x to y = a, along vertical strip Limits of x : x = 0 to x = a 3. The region is bounded by the lines y = x, y = a and x = 0 4. The point of intersection of y = x and y = a is Q (a, a). 5. To change the order of integration, i.e. to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = a Hence, the given integral after change of order is a

(

0 a

a

P

B′

y=a

Q a, a)

A′ O

x

Fig. 9.47 y y=a

P

Q (a a)

) A

y

x3 3

y

B

2

y x dy 0 3

y 3

y3

y

x

O

Fig. 9.48

9.3

∫

a

Change of Order of Integration

9.43

4 3 y y 3

4 y4 3 4

a

0

4

a 3

Example 2 Change the order of integration and evaluate

∫ ∫

sin y y

p

B¢

Solution y

1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y = p Limits of x : x = 0 to x = p 3. The region is bounded by the line y = x, y = p and x = 0. 4. The point of intersection of the line y = x and the line y = p is P (p, p). 5. To change the order of integration, i.e., O to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = p Hence, the given integral after change of order is y =∫

∫

A¢

x

Fig. 9.49

Q

A

y=p P (p p )

B

sin y y

− cos y 0 cos 2

P (p p )

y

y

sin y y x0 y y sin y y y y

y

O

x

cos 0 Fig. 9.50

9.44

Chapter 9

Multiple Integrals

Example 3 Evaluate the iterated integral

1 1

Ú0 Úx sin y

2

[Winter 2013]

dydx.

Solution 1. Since the inner limits depend on x, the function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y = 1, along vertical strip Limits of x : x = 0 to x = 1 3. The region is bounded by the lines y = x, y = 1 and x = 0. 4. The point of intersection of the line y = x and the line y = 1 is P (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1

Fig. 9.51

Hence, the given integral after change of order is

Ú0 Ú

n

y

2

=Ú

1

1

s y 2 dx y Fig. 9.52

y

n y 2 x 0 dy 2

y dy

1 1 2 ◊ y dy 2 1 1 = - co y 2 y 0 2 1 = 2 1 = 1] 2

sin ( ) ◊

( dy

- cos ( y)

9.3

Change of Order of Integration

9.45

Example 4 Change the order of integration and evaluate

∫ ∫

x

e− y y

x [Winter 2015]

y

Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = x to y Æ •, along vertical strip Limits of x : x = 0 to x Æ • 3. The region is bounded by the lines y = x and x = 0. 4. Here, the only point of intersection is origin. 5. To change the order of integration, i.e. to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = • Hence, the given integral after change of order is

∫

∞

∞

x

y

∫ {∫ ∞

∫

∞

y

x0

=

⋅

∫

= − (e 1

x

e y

e

e

y −y

y

y

e

= −e

dx y

y y

A x

O

Fig. 9.53 y

A

B

y

e

0

B′ y Æ •

y y 0

e )

x

O

y

Fig. 9.54

9.46

Chapter 9

Multiple Integrals

Example 5 Evaluate the integral

2 1 x2

Ú0 Ú y e

dx dy by changing the order of integration.

2

[Summer 2017, 2015]

Solution 1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. y 2. Limits of x : x = to x = 1 2 along horizontal strip A¢B¢ Limits of y : y = 0 to y = 2 3. The region is bounded by the lines y = 2x, x = 1, y = 2, and y = 0. 4. The point of intersection of y = 2x and x = 1 is x = 1, y = 2. The point of intersection is Q (1, 2). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the line y = 2x. Limits of y : y = 0 to y = 2x Limits of x : x = 0 to x = 1 Hence, the given integral after change of order is 2 1 x2

Ú0 Ú y e 2

dx dy = Ú

1 2 x x2

Ú

e dy dx

0 0

=Ú

1

0

{Ú } 2x

0

1

Fig. 9.55

Fig. 9.56

2

dy e x dx 2

= Ú 2 x ◊ e x dx 0

= ex

2 1

0

= e1 - e0 = e -1

È∵ e f ( x ) f ¢( x ) dx = e f ( x ) ˘ Î Ú ˚

9.3

Change of Order of Integration

9.47

Example 6

∫ ∫

Change the order of integration and evaluate

x

−

xe

x2 y

Solution 1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = 0 to y = x. Limits of x : x = 0 to x Æ •. 3. The region is the part of the first quadrant bounded between the lines y = x and y = 0. 4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip parallel to x-axis which starts from the line y = x and extends up to infinity. Limits of x : x = y to x Æ • Limits of y : y = 0 to y Æ • Hence, the given integral after change of order is

∫

x

−

xe

x2 y

∫

x =∫

−

xe

∞

=−

y 2

x2 y

∫

1 y e 2∫

1 2 1 2

B′

A

O

Fig. 9.57

dx y

y

x y

e

x y

−

2x y

y

y y

f ′ (x x

ef

x)

x

O −y

dy Fig. 9.58

y

x

y

∵∫ e 1 ∞ 2∫

y

y 0

Example 7 a

Change the order of integration and evaluate

∫∫

a y

x2 x2

y2

dx y

9.48

Chapter 9

Multiple Integrals

Solution

2

a

x2

y2

B

x2

=

x2 a

∫ ∫

dx

y2 1

x

x2

y2

O

2

dy 

2

dx

0 a

− og x x 2 dx

a

(

) 

x 1+



x

=

+

)∫

=

+

) x3

+

)

a

0

x2 x

3 a

=

a3 3

0

2

A

Fig. 9.60

x

a

= ∫ log

a

=a

1. Since inner limits depend on y, the function is integrated first w.r.t. x but evaluation becomes easier by changing the order of intey gration. 2. Limits of x : x = y to x = a, along horizontal Q (a, a) strip A¢B¢ Limits of y : y = 0 to y = a 3. The region is bounded by the lines y = x, x = a and y = 0. A′ B′ 4. The point of intersection of y = x and x = a is x = a, y = a. P x O The point of intersection is Q (a, a). 5. To change the order of integration, i.e. to integrate first w.r.t. y, draw a vertical strip AB Fig. 9.59 parallel to y-axis which starts from x-axis and y terminates on the line y = x. Q a a) Limits of y : y = 0 to y = x Limits of x : x = 0 to x = a Hence, the given integral after change of order is

P

x

9.3

Change of Order of Integration

9.49

Example 8 [Winter 2016]

Change the order of integration and evaluate 1

∫∫

ey

1− x 2

((ee + +11) 1 − x − y y

0 0

2

2

y

dy dx.

Q (0, 1) B′

Solution

P (1, 0)

1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration. 2. Limits of y : y = 0 to y = 1 − x 2

to

x = 1− y

Limits of y : y = 0

to

y=1

∫∫ 0

0

ey

1− x 2

((ee + +11) 1 − x − y y

2

2

x

Fig. 9.61 y Q (0, 1) x2 + y2 = 1 A

B

O

P(1, 0)

2

Hence, the given integral after change of order is 1

A′

O

Limits of x : x = 0 to x = 1 3. Since given limits of x and y are positive, the region is the part of circle x2 + y2 = 1 in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from y axis and terminates on the circle x2 + y2 = 1. Limits of x : x = 0

x2 + y2 = 1

1− y 2 ey 0 e y + 1 ∫0

dy dx = ∫

Fig. 9.62

1

1

((1 − y 2 ) − x 2

x e y sin −1 =∫ y 0 e +1 1 − y2

dx d y

1− y 2

1

dyy 0

ey − sin −1 0)dy (sin −−11 1 1− dy 0 e +1 y 1 e π =∫ y ⋅ dy dy 0 e +1 2 f ′( y) π 1   dy = log f ( y )  = log(e y + 1) 0 ∵ ∫ f ( y ) dy 2   =∫

1

y

π [log( og(e + +1 1) − log log 2] 2 π  e + 1 = log   2  2 =

x

9.50

Chapter 9

Multiple Integrals

Example 9 a

∫∫

Change the order of integration and evaluate Solution 1. Since inner limits depend on y, the function is integrated first w.r.t. x. 2

2. Limits of x x x a + a2 A¢B¢

a

a − y2 2

y2

d

y P (a a)

y 2 to

A′

B

y 2 , along horizontal strip

Limits of y : y = 0 to y = a 3. The region is bounded by the circle (x – a)2 + y2 = a2 and the line y = 0. Since limits of y are positive, the region is the part of the circle (x – a)2 + y2 = a2 above x-axis. 4. The points of intersection of the circle with x-axis are O (0, 0) and Q (2a, 0). 5. To change the order of integration i.e. to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the circle

Q (2a, 0) O

x

a, 0)

Fig. 9.63 y B

(x – a)2 + y2 = a2 or x2 + y2 – 2ax = 0 Limits of y : y = 0

Q (2a, 0) 2

to

O

A

x

(a

Limits of x : x = 0 to x = 2a Hence, the given integral after change of order is

∫

a

y

d

x2

y

∫

2a

=

2a

=

2a

=

2

y

− x2

0

− x2 x 2

− x a)2

( x a) 2 a 2

a = 2

Fig. 9.64

x a)2

a2 (0 a + n 11 2 2

a2 sin − 2 a2 0− 2

x a a

2a

0

( −1)

9.3

Change of Order of Integration

9.51

-1 = a 2 sin -1 1 [∵ sin -1 ( -1) = - sin -1 (1)]

p 2 p a2 = 2 = a2

Example 10 1 2- x2

Evaluate

Ú Ú 0

x

x x 2 + y2

dy dx by changing the order of integration. [Summer 2016]

Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y. 2. Limits of y : y = x to y = 2 - x 2 along vertical setup A¢B¢ Limits of x : x = 0 to x = 1 3. The region is bounded by y-axis, the line y = x and the circle x2 + y2 = 2. 4. The point of intersection of the circle y = 2 - x 2 and y = x is obtained as x2 = 2 - x2 2 x2 = 2

\

x2 = 1 x = ±1 y=1

Fig. 9.65

Hence, P(1, 1) is the point of intersection. 5. To change the order of integration, i.e. to integrate first w.r.t. x, divide the region into two subregions OPR and PQR. Draw a horizontal strip parallel to x-axis in each subregion. (i) In the subregion OPR, strip AB starts from y-axis and terminates on the line y = x. Limit of x: x = 0 to x = y Limit of y: y = 0 to y = 1

Fig. 9.66

9.52

Chapter 9

Multiple Integrals

(ii) In the subregion PQR, strip CD starts from y-axis and terminates on the circle y = 2 - x2 . Limit of x: x = 0 to x = 2 - y 2 Limit of y: y = 1 to y = 2 1 2 - x2

Ú Ú 0

x

x x +y 2

2

dy dx =

x

ÚÚ

x +y 2

OPR 1 y

x +y 2

00

x

ÚÚ

x + y2 2

PQR

2 2 2- y

x

= ÚÚ

dy dx +

2

2

dy dx +

1 È 1 1 Í ( x 2 + y2 ) 2 =Ú Í 1 2Í 0 2 ÎÍ 1

0

y2 = ( 2 - 1) 2

1

2

Ú ÈÎ 1

x + y2 2

0

y

˘ ˙ ˙ dy + ˙ ˚˙ 0

= Ú ÈÎ 2 y - y ˘˚ dy +

x

Ú Ú 1

2

Ú 1

2 - y ˘˚ dy

È y2 ˘ + Í 2y - ˙ 2 ˙˚ Í 1 0 Î

2 1 3 - + - 2 2 2 2

= 1- 2 + = 1-

1 2

1 2

dy dx

1 È 1 Í ( x 2 + y2 ) 2 Í 1 2Í 2 ÎÍ

2

1ˆ Ê1 ˆ Ê = ( 2 - 1) Á - 0˜ + Á 2 - 1 - 2 + ˜ Ë2 ¯ Ë 2¯ =

dydx

˘ ˙ ˙ ˙ ˚˙ 0

2 - y2

dy

9.3

Change of Order of Integration

9.53

Example 11 Sketch the region of integration, reverse the order of integration and 2y 2 4 - x 2 xe dy dx. evaluate Ú0 Ú0 [Summer 2014] 4-y Solution 1. Since inner limit depends on x, the function is integrated first w.r.t. y. 2. Limits of y : y = 0 to y = 4 – x2 along vertical strip A¢B¢ Limits of x : x = 0 to x = 2 3. The region is bounded by the parabola x2 = 4 – y, x-axis and y-axis. 4. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip parallel to x-axis which starts from y-axis and terminates on the parabola x2 = 4 – y. Limits of x : x = 0

to

x = 4− y

Limits of y : y = 0

to

y=4

Fig. 9.67

Fig. 9.68

Hence, the given integral after change of order is 2

4 - x2

Ú0 Ú0

2y 4 4 - y xe xe2 y dydx = Ú Ú dx dy 0 0 4-y 4-y

=Ú

4

=Ú

4

0

0

=Ú

4

0

e2 y È 4 - y ˘ x dx ˙ dy 4 - y ÍÎ Ú0 ˚ e2 y x 2 4-y 2

4- y

dy 0

e2 y È 1 ˘ ( 4 - y ) ˙ dy Í 4 - y Î2 ˚

9.54

Chapter 9

Multiple Integrals

=Ú

4

0

e2 y dy 2

1 2y 4 e 0 4 1 = (e8 - 1) 4 =

Example 12 Change the order of integration and evaluate

1

2− x

0

x2

∫∫

xy dx dy.

Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y.

P ′(−2, 4) y

The correct form of integral =∫

1

0

∫

2− x x2

xxyy dy dx

Q (0, 2) 2

2. Limits of y : y = x to y = 2 – x, along vertical strip A¢B¢ Limits of x : x = 0 to x = 1 3. The region is bounded by y-axis, the line x + y = 2 and the parabola x2 = y. Since given limits of x and y are positive, the region lies in the first quadrant. 4. The points of intersection of x + y = 2 and x2 = y are obtained as x2 = 2 – x x +x–2=0 (x – 1) (x + 2) = 0 x = 1, –2 y = 1, 4

x2 = y

B′ A′

P (1, 1) x +y=2

O

x

Fig. 9.69 P ′(−2, 4) y

2

Q (0, 2) C R A

D B

x2 = y

P (1, 1) x +y=2

O x The points of intersection are P (1, 1) and P' (–2, 4). Fig. 9.70 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into two subregions OPR and RPQ. Draw a horizontal strip parallel to x-axis in each subregion. (i) In subregion OPR, strip AB starts from y-axis and terminates on the parabola x2 = y. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 1

9.3

Change of Order of Integration

9.55

(ii) In subregion RPQ, strip CD starts from y-axis and terminates on the line x + y = 2. Limits of x : x = 0 to x = 2 – y Limits of y : y = 1 to y = 2 Hence, the given integral after change of order is

∫∫

x

y

x

∫

dx 1 0

x2 2

y

+∫

x2

2

2 y

2 y

y y

1

0

1 2 1 = 2

xy dx y

0

2

=

y

1

+

2 1

1

1 3 2 3 0 2 1 1 1 + 2 3 2 1 5 = + 6 24 9 24 3 = 8

4

2

2 32 − 3

y 4 2

3

3

4

1

4 1 + − 3 4

Example 13 Change the order of integration and evaluate

4a

∫ ∫

2 ax x2 4a

Solution 1. Since inner limits depend on x, the function is integrated first w.r.t. y. x2 2. Limits of y y to x 4a along vertical strip A¢B¢ 3. The region is bounded by the parabolas x2 = 4ay and y2 = 4ax. 4. The points of intersection of x2 = 4ay and y2 = 4ax are obtained as x4 = 16a2y2 =16a2 (4ax)

x. [Winter 2014]

y y 2 = 4ax

P(4a, 4a)

B′ x 2 = 4ay A′ O

Fig. 9.71

x

9.56

Chapter 9

Multiple Integrals

x(x3 – 64a3) = 0 x = 0, x = 4a \ y = 0, y = 4a The points of intersection are O (0, 0) and P (4a, 4a). 5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts from the parabola y2 = 4ax and terminates on the parabola x2 = 4ay. y

y2 to x = 2 a ay Limits of x : x = 4a Limits of y : y = 0 to y = 4a Hence, the given integral after change of order is 4a

2 ax

Ú0 Ú x

2

4a

xy dy dx = Ú xy

0

4a

2 ay

Úy

2

0

=Ú

x

P(4a, 4a) x 2 = 4ay A

dx dy

4a 4a 2 ay

=Ú

y 2 = 4ax

y2 4a

O

y2 ˆ 2 ay dy Á 4 a ˜¯ Ë

=2 a

3 y2

3 2

x

dy

4a Ê

0

B

Fig. 9.72

4a

1 y3 4 3 4a

4a

0

0

3

3

4 1 a ( 4) 2 a 2 (64 a 3 ) 3 12 a 32 2 16 2 = a - a 3 3 16 2 = a 3 =

Example 14 Change the order of integration and evaluate a

a− a − a2 − y2

0

0

∫∫

xy log( x + a ) dx dy. ( x − a)2

Solution 1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration.

9.3

2. Limits of x : x = 0 to x = a − a 2 − y 2

y y=a

Limits of y : y = 0 to y = a 3. The region is bounded by the circle (x – a)2 + y2 = a2, the lines y = a and x = 0.

Q A′

4. The point of intersection of (x – a)2 + y 2 = a2 and y = a is obtained as (x – a)2 + a2 = a2 x=a The point of intersection is P (a, a). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from the circle (x – a)2 + y2 = a2 and terminates on the line y = a. axx − x 2 Limits of y : y = 2a

to

Limits of x : x = 0

to x = a

P (a, a)

B′

(x − a)2 + y 2 = a 2 x

O

Fig. 9.73 y y=a Q

B

P((a, a) (x − a)2 + y 2 = a 2

A

y=a x

O

Hence, the given integral after change of order is 2

a

a− a − y

0

0

∫∫

2

9.57

Change of Order of Integration

Fig. 9.74

a a xy log( x + a ) x log( x + a ) dx d y = ∫ ∫ y dy d x 2 2 0 2 ax − x ( x − a) ( x − a)2

=∫

a

=∫

a

0

0

x log( x + a ) y 2 2 ( x − a)2

a

dx dx 2 ax − x 2

x log( x + a )  a 2 − 2aaxx + x 2  dx  dx 2 ( x − a ) 2 

1 a x log( x + a) a )dx 2 ∫0 a 2  a x 1  x2 1 =  log( x + a ) − ∫ ⋅ dx  dx 0 2 2 2 x+a  0   =

=

1  a2 1 a a2   dx   dx  log 2a − ∫0 ( x − a ) + 2 2 2  x + a 

a  1  a2 1 x2  log 2a − − ax + a 2 log( x + a )  2 2 2 2 0   2   1 a =  a 2 log 2a − + a 2 − a 2 log 2a + a 2 log a 4 2 

=

9.58

Chapter 9

Multiple Integrals

=

 1  a2 + a 2 log a  4 2 

=

a2 (1 + 2 log a ) 8

Example 15 Change the order of integration and evaluate 1 2

∫ ∫ 0

1 + x2

1 4 y2 1−

0

1 − x2 1 − x2 − y 2

dx dy. y y

Solution Q 0,

1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. 2. Limits of x : x = 0

O

1 2 3. Since the limits of x and y are positive, the region is the part of the ellipse in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from x-axis and terminates on the ellipse x2 + 4y2 = 1. Limits of y : y = 0

x 2+ 4y 2 = 1 B′ x P (1, 0)

A′

x = 1 − 4 y2

to

1 2

y=

to

Fig. 9.75 y Q 0,

1 2

B x 2+ 4y 2 = 1

O

Limits of y : y = 0 Limits of x : x = 0

to to

Ú Ú

P (1, 0)

x=1

Hence, the given integral after change of order is 1 1- 4 y2 2 0 0

A

1 y= 1 − x2 2

1 + x2 1 - x 2 1 - x 2 - y2

Fig. 9.76

dxdy = Ú

1

=Ú

1

0

0

1 + x2 1- x

2 Ú

1 + x2 1- x

1 1- x 2 2 0

2

sin -1

1 (1 - x 2 ) - y 2 y 1 - x2

dydx

1 1- x 2 2

dx 0

x

9.3

1 + x2

=Ú

1

=Ú

12

= =

1- x

p 6

Ú

sin -

2

2

s

-1

9.59

0 dx

1- x ) p ◊ d 6 1 x2

1

1 x

p 6

-

1 x2

2

-

2

x -

x 1 2

-1

∵ Ú a2 - x2 d

=

Change of Order of Integration

1

x 0

x 2

2

x2 +

a2 2

n

x a

p 3 -1 sin 1 6 2

p p ◊ 4 2 p2 = 8 =

Example 16 Change the order of integration and evaluate a

∫∫

x

y

(a

)(

dx y )( y

x)

Solution y

1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. 2. Limits of x : x = 0 to x = y Limits of y : y = 0 to y = a 3. The region is bounded by the line y = x, y = a and x = 0. 4. The point of intersection of y = a and y = x is P (a, a). 5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis which starts from the line y = x and terminates on the line y = a.

y=a P (a, a)

A′

B′

x

O

Fig. 9.77

9.60

Chapter 9

Multiple Integrals

Limits of y : y = x to y = a Limits of x : x = 0 to x = a Hence, the given integral after change of order is a y x dx

∫∫

(a =∫ =∫

a

)(

∫

y )( y

(a a2

x2

∫

P (a, a)

dx

x )(

a

=a

B

x)

x

a

y

y )( y

x)

A



y

a x

(a

)(

x)

dx

x

O

Putting y – x = t2, dy = 2t dt When y = x, t = 0 When

∫

Fig. 9.78

x

y = a,

a

0

a

x (

y y

2t t

a x

= x

a = 2∫

a

= 2∫

a

0

x x a

2

x

2

x a2

0

(

=2

a

= ⋅

2∫

x2

∫

sin −1

x  1 ( 2 -

a−x

t a x

dx

dx 0

− sin 0)dx

a − x2

p - 2( 2

x) t 2

(

dx

2

dt

a x

2

a

x t

−

1

x ) 2 ( 2x) d 1 )2

a

È ∵Ú

0

( x )]n f ( )dx =

[ ( x )]n +1 ˘ n +1

pa

Example 17 Change the order of integration and evaluate 1

∫∫

1 x x

y (

x

y )

Solution 1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing the order of integration.

9.3

y

1 x

2. Limits of y : y = x

to

y

Limits of x : x = 0

to

x=1

xy

3. The region is bounded by the rectangular hyperbola xy = 1, the line y = x and y-axis in the first quadrant. 4. The point of intersection of xy = 1 and y = x in the first quadrant is obtained as x2 = 1 x=1 \ y=1 The point of intersection is P (1, 1). 5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region into two subregions OPQ and QPR. Draw a horizontal strip parallel to x-axis in each subregion. (i) In subregion OPQ, strip AB starts from y-axis and terminates on the line y = x. Limits of x : x = 0 to Limits of y : y = 0 to

9.61

Change of Order of Integration

1

B′ (1, 1) A′ O

x

Fig. 9.79 y

R

D

C

xy

x=y y=1

1 (1, 1)

Q

(ii) In subregion QPR, strip CD starts from y-axis and terminates on the rectangular hyperbola xy = 1. 1 Limits of x : x = 0 to x y Limits of y : y = 1 to y → ∞

O

x

Fig. 9.80

Hence, the given integral after change of order is 1

∫∫

1 x

y

∫

x=

y

y 1+

1

y

∫

+

− 1



1 1+ y

−

1  +y

y 1+

1 2

1+ y

1

∞ 1

1 1+ y

1

∞

y

xy ) 0

1 1 y 2 1 y

d

y

y

1 y

1

∞

0

=− = −∫

1 y

dx y +

y

9.62

Chapter 9

Multiple Integrals

Putting y = tan q in the first term of first integral, dy = sec2q dq, When y = 0, q = 0 When y = 1, θ = 1

1 x

0

x

∫∫

π 4

π 1 ∞ y sec 2 θ dθ 1 4 d y d x = − + ta tan n −1 y 0 + tan −1 y 1 2 2 4 ∫ 0 2 ((1 1+ 1 + xy ) (1 + y ) sec θ π

= −∫ 4 0

(1 1+ + cos 2θ ) 1 dθ + (tan −−11 1 − tan −1 0) + (tan −−11 ∞ − tan −1 1) 2 2

1 sin 2θ =− θ+ 2 2

π 4

+

0

π 1π π +  −  4 2  2 4

π 1 π 3π = − − sin + 8 4 2 8 π −1 = 4

Example 18 Change the order of integration and evaluate 1

∫∫ 0

1− y 2

0

cos −1 x 1 − x2 1 − x2 − y 2

dx dy.

Solution y

1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing the order of integration. 2. Limits of x : x = 0 to x = 1 − y Limits of y : y = 0 to y = 1 3. Since given limits of x and y are positive, the region is the part of the circle x2 + y 2 = 1 in the first quadrant. 4. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB parallel to y-axis in the region. AB starts from x-axis and terminates on the circle x2 + y 2 = 1. Limits of y : y = 0 to y = 1 − x 2 Limits of x : x = 0 to x = 1.

Q (0, 1)

2

A′ O

Fig. 9.81

B′

x2 + y2 = 1

P (1, 0) x

9.3

Change of Order of Integration

Hence, the given integral after change of order is 1

∫∫

1 y

9.63

y

1

cos x

2

− x2

1 =∫

1

=∫

1

=∫

1

∫

y2

cos 1 x

1− x 2

cos 1 x

=− −

cos x 2∫

1

2

B

)

y

2

x2 + y2

x

sin

− x2

(sin −1

A

O

1 x2 −

1 1

Q

1

2

1

1− x

dx y

1

P (1, 0) x

dx 0

n − )dx Fig. 9.82

cos −1 x −

(cos 1 x 2 2

−

π (cos 4

−

π π 0− 4 2

2 1

0

1

1 1 − x2

dx ( x)]2  2

∵ ∫ f ( ) f ( )d x = 2

2

3

16

EXERCISE 9.3 Change the order of integration of the following integrals: 1.

6

∫∫

2+ x

2 x

y y dx 

2.

1

∫∫

2x

x

y y dx 

3.

1

∫∫

y y



6

∫

Ans. :

0

∫

y

1

y 2

2



y x dy Ans.

1

∫ ∫

1 2

y x dy 

9.64

4.

Chapter 9

y a

a

∫ ∫

Multiple Integrals

y x dy

a

a

Ans. 5.

a

2a x

∫∫

3

∫ ∫

y

2 y

ax

y y dx

x2 a



6.

:

y x dy

6

1

y

∫∫

2

∫∫

6 x 2 x2 +4 4

2

∫∫

x

a

a

∫ ∫

y +a a2 − y 2

0

0

∫

2a

∫ ∫

1

x x2





∫

2 4

6 +2

y

2

f (x , y )dxdy

2 ax

a

a

∫



a x a

y x dy

2 ax − x



∫

2

Ans.

11.

x 2

y y dx

4 x

 Ans. : 10.

∫

:

x

Ans. :

9.



x

y

y



8.

x

x

Ans. :

7.

a

: a

∫

2

a 2a

2

2a

∫ ∫

2a y 2a

f x

y

9.3

12.

a

∫∫

y y dx a 2

a2 x

a

∫∫

b

∫ ∫

mx

1

∫∫

e

2

Ans. :

k a k b

∫

k y

ma

b

a

y m

17.

1 2

y y dx

x 2

∫∫

a

8

1

∫∫

gy

2

y

y x y

1



y3

Ans. : ∫

1

∫

1 x2 2

2   3

1+ x 2 1 x

y

x 1 

19.

∫

1

2

2

Ans.

e

1+ x 2

1 4y2

∫ ∫





18.



y y dx

x3

2

∫∫

y

x

Ans. : ∫ 16.



a2

y

y y dx

k x

  Ans. :  15.



∫

y y dx

x



14.

9.65

4

Ans.

13.

Change of Order of Integration

∫∫

a

Ans.

2

∫∫

2 2y

x 1

1 4

4)

x

y

a 4

9.66

Chapter 9

9.4

DOUBLE INTEGRALS IN POLAR COORDINATES θ2

The integral

∫∫

θ1

r2 r1

Multiple Integrals

f ( r , θ )drdθ represents the polar form of the double integration. This

integral is first integrated w.r.t. r keeping q constant and then the resulting expression is integrated w.r.t. q. R

Limits of Integration If the limits of integration are not given then these limits are obtained from the equations of the given curves. Let the region be bounded by the curves r = r1(q ), r = r2 (q ) and the lines q = q1, q = q2.

B = q2

r

r (q )

Q S A

= q1

dq

The region of integration is PQRS. Draw an elementary radius vector AB from origin which enters in the region from the curve r = r1(q ) and leaves at the curve r = r2(q ). Therefore, limits for r are r1(q ) to r2(q ).

q2

P

r

q O

r (q ) =0

Fig. 9.83

To cover the entire region PQRS, rotate elementary radius vector AB from PQ to RS. Therefore, q varies from q1 to q 2.

∫∫ f

r,

(

r 1

( )

f (r

r

Type I Evaluation of Double Integrals in Polar Coordinates

Example 1 Evaluate

∫ ∫ 4

1

Solution

∫ ∫ 4

0

1

r rd

r r d 1

∫

4

0

∫

4

0

1 2

r2 d 2 0 1 d 2 4 0

9.4

Double Integrals in Polar Coordinates

9.67

1 π ⋅ 2 4 π = 8 =

Another method: Since both the limits are constant and integrand (function) is explicit in r and q, the integral can be written as π 4 0

∫ ∫

1

0

π

1

r dr dθ = ∫ 4 dθ ⋅ ∫ r dr 0

0

π

= θ 04 ⋅

r2 2

1

0

π 1 ⋅ 4 2 π = 8 =

Example 2 Evaluate

π

sin θ

0

0

∫ ∫

r dr dθ.

Solution π

sin θ

0

0

∫ ∫

π sin θ r dr dθ = ∫  ∫ r dr  dθ  0  0 

=∫

π

0

r2 2

sin θ

dθ 0

1 π 2 sin θ dθ 2 ∫0 1 π 1 − cos 2θ = ∫ dθ 2 0 2

=

=

1 sin 2θ θ− 4 2

π

0

1 sin 2π  π −  4 2  π = 4 =

9.68

Chapter 9

Multiple Integrals

Example 3 Evaluate the integral

p 2 1 - sin q r 2 0 0

Ú Ú

cos q dr dq .

[Summer 2014]

Solution p 1 - sin q 2 2 r 0 0

Ú Ú

cos q dr dq =

=

=

p 2 0

È

r dr ˘˙ cos q dq ˚

1 - sin q 2

Ú ÍÎÚ0 p 2 0

Ú

p 2 0

1 3

1 - sin q

r3 3

Ú

cos q dq 0

(1 - sin q )3 cos q dq

1 (1 - sin q )4 = 3 4 =

p 2 0

1˘ È Í0 - 4 ˙ Î ˚

1 3

1 12

= -

Example 4 π 2 a cos θ 2 0 0

∫ ∫

Evaluate

r 2 sin θ dθ dr.

Solution Since inner limits depend on q, the function is integrated first w.r.t. r. π

2 a cos θ

The correct form of the integral = ∫02 ∫0 π 2 a cos θ 2 0 0

∫ ∫

π

2 a cos θ

r 2 sin θ dr dθ = ∫ 2  ∫ 0   0 π 2 0

=∫

r 2 sin θ dr dθ

r3 3

r 2 dr  ssin in θ dθ 

2 a cco os θ

sin θ dθ 0

π

=

8a 3 2 coss3 θ sin sin θ dθ 3 ∫0 π

=−

8a 3 2 coss3 θ ( − sin sin θ ) dθ 3 ∫0

9.4

8a 3 =3

4

p 2

q

4

Double Integrals in Polar Coordinates

∵Ú f

¢ )dq

0

9.69

˘ f (q )]n +1 , n π -1 n +1

3

==

8a 12

s4

- cos 4 0

2

2 3 a 3

Example 5 Evaluate

∫ ∫ 2

a (1 sin )

2

Solution Since inner limits depend on q, the function is integrated first w.r.t. r.

∫ ∫ 2

a

r

2

0

a

r

∫

sin )

3

cos 0

p

+s

0

a3 3

s 4

a3 12

+ sin

a3 12 5 = a3 4

−1

Example 6

∫ ∫ 4

cos 2

(

r r

r

3 a (1 sin )

=

Evaluate

r2

r 2 r cos

2

1 3

=

(1+ sin )

∫ ∫

The correct form of the integral

2

dr d

cos q dq p 4 2

∵Ú

[ ( )]n +1 , n +1 n π -1

) dq

f

0

π

4

−

+

0) 4

9.70

Chapter 9

Multiple Integrals

Solution p cos 2q 4 0 0

Ú Ú

1 2r ◊ dr dq 2 (1 + r 2 )2 p

1 4 È cos 2q ˘ (1 + r 2 )-2 ◊ 2r dr ˙ dq Ú Ú Í 0 0 Î ˚ 2 n +1 È f (r )] ˘ p [ n cos 2q 1 4 Í ˙ dq Í∵ Ú [ f (r )] f ¢(r ) dr = n + 1 ˙ = Ú -(1 + r 2 )-1 0 2 0 Í ˙ n π -1 Î ˚ =

p

=-

1 4Ê 1 ˆ - 1 dq 2 Ú0 ÁË 1 + cos 2q ˜¯

=-

1 4Ê 1 ˆ - 1 dq 2 Ú0 ÁË 2 cos2 q ˜¯

=-

1 4Ê1 ˆ sec 2 q - 1˜ dq Á Ú 0 Ë ¯ 2 2

p

p

p 4 1 1 =tan q - q 2 2 0

1Ê1 p pˆ = - Á tan - ˜ 2Ë2 4 4¯ =

1 (p - 2) 8

Example 7 Evaluate

p p 2 2 0 0

Ú Ú

sin(q + f )dq df.

Solution p p 2 2 0 0

Ú Ú

p È p ˘ sin(q + f) f ) dq df = Ú 2 Í Ú 2 ssin((q + f ) dq ˙ df 0 ÍÎ 0 ˙˚ p 2

p 2 0

=Ú

- cos(q + f ) df 0

p 2 0

= -Ú

p

È Êp ˘ ˆ Ícoss ÁË + f ˜¯ - cos f ˙ df 2 Î ˚

= - Ú 2 ( - sin f - cos f ) df 0

9.4

Double Integrals in Polar Coordinates

= - coss f - ssin in f

9.71

p 2 0

p p Ê ˆ = - Á coss - sin - coss 0 + ssin in 0˜ Ë ¯ 2 2 =2

Type II Evaluation of Double Integrals Over a Given Region in Polar Coordinates

Example 1 Evaluate

ÚÚr

3

sin 2q dr dq over the area bounded in the first quadrant

R

between the circle r = 2 and r = 4.

[Summer 2016]

Solution 1. The region of integration is the interior of the circle between r = 2 to r = 4. 2. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = 2 and leaves at the circle r = 4. 3. Limits of r: r = 2 to r = 4 Limits of q: q = 0 to q =

p 2

I = ÚÚ r 3 sin 2q dr dq p 2 4

=

Ú Úr

3

sin 2q dr dq

0 2

r4 = 4

4

2

cos 2q 2

p 2 0

=

1È 4 Ê 1ˆ (4) - (2)4 ˘˚ Á ˜ (cos p - cos 0) Î Ë 2¯ 2

=

1 Ê 1ˆ [256 - 16] Á - ˜ (-2) Ë 2¯ 2

1 [240] 2 = 120 =

Fig. 9.84

9.72

Chapter 9

Multiple Integrals

Example 2 a2

ÚÚ r

Evaluate

2

over the upper half of the circle

q

r = a cos q.

[Summer 2017]

Solution

q

p 2

1. The region of integration is the upper half of the circle r = a cos q. 2. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = a cosq. 3. Limits of r : r = 0

to

A r = a cos q

r = a cosq O

=

Limits of q : q = 0 to 2

I p

2

r 2 dr dq 1 2 ) (-2r

1 - (a 2

q

a

q =0

3

a

r dq

q

p 2

-

1 2Ú

-

1 2 ( 3Ú

s

-

a 3

p 2

3

Fig. 9.85

[ (r )]

)dr

p

Ú

-a ) q 3q 4

cos 3q + 3

a 1 4 3

- 1 dq

q

p 2 0

p p p 3 + cos - + co 2 12 2 2 4

=-

a 3

3 4

=-

a3 3

p 3 + 2 4

=-

a3 2 p 3 3 2

1 12

-

1 cos 0 12

(r )]n +1 n 1

9.4

9.73

Double Integrals in Polar Coordinates

Example 3 4

ÚÚ

Evaluate

3

over the interior of the circle r = 2a cos q.

Solution 1. The region of integration is the interior of a cos the circle r

q

p A

2. Draw an elementary radius vector OA which starts from the origin and termia cos nates on the circle r 3. Limits of r r

r = 2 cos q

a cos O

=−

Limits of

to

2

q

0

2

r 4 cos3 q r q

I =Ú

p 2 p 2

=Ú

p 2 p 2

Ú

2 a cosq 4

r5 5

r

3

Fig. 9.86

r q

q

2a

s3 q q 0

p

=

1 2 (2 cos q 5Ú p

cos q dq

2

p

=

32 a 5 ◊ 2 Ú 2 cos8 0 5

=

64 a 5 7 5 3 1 p ◊ 5 8 6 4 2 2

cos8

= co

8

[Using reduction formula]

7p 5 a 4

Example 4 Evaluate

ÚÚ r

initial line.

2

r q over the cardioid r = a (1 + cos q ) above the

9.74

Chapter 9

Multiple Integrals

Solution 1. The region of integration is the part of the cardioid r = a (1 + cos q ) above the initial line (q = 0). 2. Draw an elementary radius vector OA which starts from the origin and terminates on the cardioid r = a (1 + cos q ). 3. Limits of r : r = 0 to r = a (1 + cos q ) Limits of q : q = 0

q=p 2

to q = p

r dq

I

Ú Ú Ú

r3 3

p

1 3

a3 3

=-

a3 3

Ú

r2 n

O

q =0

nq d 0

q ◊ dq

cos p

(1 + cos q )

r dq

a 1+ cosq )

p

=-

r

a (1+ cosq

p

A

(1 + + cos 4

Fig. 9.87

- n )dq p

4

∵Ú f

)dq

0

(q )]n +1 ˘ n +1

3

=-

a ( + 12

=-

a3 12

=

-

4

4 2 a 3

Example 5 Evaluate

ÚÚ

q r

2

a

2

over one loop of the lemniscate r2 = a2 cos 2q.

Solution 1. The region of integration is one loop of the lemniscate r2 = a2 cos 2q bounded between the lines

=−

4

and

4

.

9.4

Double Integrals in Polar Coordinates

2. Draw an elementary radius vector OA which starts from the origin and terminates on the lemniscate r 2 = a2 cos 2q.

q= p 2 q= p 4 r 2 = a 2 cos 2q A q=0

3. Limits of r: r = 0 to r = a cos 2θ Limits of θ : θ = −

π π to θ = 4 4

q=− p 4

r 2 + a2

p

= Ú 4p Ú

0 4 p a cos 2q 4 p 0 4

Ú

r 2 + a2 1

1 2 (r + a 2 ) 2 ((2r )d )dr dq 2 1 a cos 2q

p

1 = Ú 4p 2((rr 2 + a 2 ) 2 2 4

=

1 2Ú

p 4 p 4

p

= a Ú 4p -

=a

Fig. 9.88

r d r dq

a cos 2q

-

=Ú

O

r dr dq

I = ÚÚ

9.75

dq 0

È ˘ [ f (r )]n +1 n , n π - 1˙ ¢ ( r ) dr = Í∵ Ú [ f (r )] f ¢( n +1 ÎÍ ˚˙

1 È ˘ 2a Í(cos 2q + 1) 1) 2 - 1˙ dq ÍÎ ˙˚

(

)

2 cos q - 1 dq

4

2 sin q - q

p 4 -

p 4

È p p Ê p ˆ Ê p ˆ˘ = a Í 2 sin - - 2 sin Á - ˜ + Á - ˜ ˙ Ë 4¯ Ë 4¯˚ 4 4 Î pˆ Ê = aÁ2- ˜ Ë 2¯ =

a (4 - p ) 2

Example 6 Evaluate

2 ÚÚ r dr dq over the area between the circles r = a sin q and

r = 2a sin q. Solution 1. The region of integration is the area bounded between the circle r = a sin q and r = 2a sin q.

9.76

Chapter 9

Multiple Integrals

2. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = a sin q and leaves at the circle r = 2a sin q. 3. Limits of r : r = a sin q to r = 2a sin q Limits of q : q = 0 to q = p I

q= p 2

B r = 2 sin q

rd

∫ ∫ ∫

2a n a n

r3 3

r2

A

d

r = a sin q

2a n

d O

a sin

1 (8a n 3 3∫ 7a3 sin 3 3 ∫ 7a3 3 3 ∫ 4

=

7a3 −3 co 12

=

7a  −3 12

+

s

q =0

d Fig. 9.89

3

co 3 3

d

0

3

−

+

1 3

0)

7 a 3 16 12 3 28 3 = a 9 =

EXERCISE 9.4 Evaluate the following integrals: r

1.

∫∫ re a co 2

in dr dq over the upper half of the circle r = 2a cos q. Ans. :

2.

∫∫ r

a2 1 3+ 4 16 e

r q over the region between the circles r = 2 sin q and r = 4 sin q. 

Ans.

45π 2

9.5

3.

4.

∫∫ r sinq dA ∫∫

r r +4 2

Change of Variables

9.77

over the cardioid r = a (1+ cos q ) above the initial line. 4 3   Ans. : 3 a   

dr dq over one loop of the lemniscate r2 = 4 cos 2q. Ans. :((4 − π )

9.5

CHANGE OF VARIABLES

9.5.1

Change of Variables from Cartesian to Polar Coordinates

The double integral can be changed from Cartesian coordinates (x, y) to polar dx = coordinates (r, q ) by putting x = r cos q, y = r sin q. Then ∫∫ f ( x, y )dy dx in θ ) J dr dθ where ∫∫ f (r cos θ , r ssin

J is the Jacobian (functional determinant)

defined as ∂x ∂x ∂ ( x , y ) ∂r ∂θ J= = ∂ ( r , θ ) ∂y ∂y ∂r ∂θ coss θ − r sin θ = sin n θ r cos θ = r (cos 2 θ + sin 2 θ ) =r

Hence,

dx = ∫∫ f (r cos θ , r sin sin θ ) r dr dθ ∫∫ f ( x, y)dy dx = ∫∫ f (r cos θ , r ssin in θ )r dr dθ

Example 1 Evaluate

∫∫

1 − x2 − y 2 dx dy over the first quadrant of the circle 1 + x2 + y 2

x2 + y 2 = 1. Solution 1. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 = 1 is obtained as r = 1. 2. The region of integration is the part of the circle r = 1 in the first quadrant.

9.78

Chapter 9

Multiple Integrals

3. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 1. 4. Limits of r : r = 0 to r = 1 π Limits of q : q = 0 to θ = 2 Hence, the polar form of the given integral is

q= p 2

A r=1 O

q =0

1− x − y dx dy 1 + x2 + y 2 2

I = ∫∫ π

= ∫2∫ 0

2

1− r2 r dr dθ 1+ r2

1

0

Putting r 2 = cos2t, 2r dr = – 2 sin 2t dt π When r = 0, t = 4 When r = 1, t = 0 π

0

I = ∫ 2 ∫π 0

π 2 0

=∫

4

1 − cos 2t ( − sin 2t dt) t )dθ 1 + cos 2t

π 4 0

2 sin 2 t sin 2t dt dθ 2 cos 2 t

∫

π

π

= ∫2∫4 0

0

π

Fig. 9.90

sin t ⋅ 2 sinn t cos cos t dt dθ cos t π

= ∫ 2 dθ ∫ 4 (1 − cos 2t )dt 0

=θ

0

π 2 0

sin 2t t− 2

π 4

dθ

0

π  π 1  −  2  4 2 π = (π − 2) 8 =

Example 2 Evaluate

4 xy

ÚÚ x 2 + y2 e

- x 2 - y2

dx dy over the region bounded by the circle

x2 + y 2 – x = 0 in the first quadrant. Solution 1. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 – x = 0 is r 2 – r cos q = 0, r = cos q.

9.5

2. The region of integration is the part of the circle r = cos q in the first quadrant. 3. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = cos q. 4. Limits of r : r = 0 to r = cos q

q

A r = cos q q =0

2

Hence, the polar form of the given integral is 2 2 4 xy I = ∫∫ 2 e x y 2 x y =∫

∫

4r 2 co sin e r2

co

co

= −2 = −2∫ co

sin e − r

Fig. 9.91

r dr d

−

) dr d

2

r

ef

r)

)d

= −2 = −∫ 2 e

=−e

r2

9.79

p 2

O

Limits of q : q = 0 to

2

Change of Variables

cos

2

=− e +

(2

+

cos 2 2

2

2 2 0

−e −

os 0 2

1 = e

Example 3 Evaluate

ÚÚ ( x

x 2 y2

over the region bounded by the circles y ) x2 + y2 = a2 and x2 + y2 = b2 (a > b). Solution 1. Putting x = r cos q, y = r sin q, polar form of (i) the circle x2 + y2 = a2 is r2 = a2, r = a. (ii) the circle x2 + y2 = b2 is r 2 = b2, r = b.

9.80

Chapter 9

Multiple Integrals

2. The region of integration is the part bounded between the circles r = a and r = b. 3. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = b and terminates on the circle r = a. 4. Limits of r : r = b to r = a Limits of q : q = 0 to q = 2p Hence, the polar form of the given integral is I

ÚÚ ( x 2p

x2 y 2 y ) a

Ú

2p

sin q r2 4

cos

B r A O

a

r=b =0

dx y

r 4 co

b

q= p 2

s

q

4

◊ r r dq Fig. 9.92

a

dq b 4

sin 2q (a ) ◊ dq 4 4 4 2 ( a 4q ) d Ú 16 2 2p

Êa

4

ˆ

32 Êa

4

32 p 16

-

ˆ

q-

n 4q 4

2p 0

(2p ) )

Example 4 Evaluate

ÚÚ

(

)2

over the region common to the circles x 2 y2 x2 + y 2 = ax and x2 + y2 = by (a, b > 0).

Solution 1. Putting x = r cos q, y = r sin q, polar form of (i) the circle x2 + y 2 = ax is r 2 = ar cos q, r = a cos q. (ii) the circle x2 + y 2 = by is r 2 = br sin q, r = b sin q. 2. The region of integration is the common part of the circles r = a cos q and r = b sin q.

9.5

Change of Variables

9.81

q= p 2

3. The point of intersection of the circle r = a cos q and r = b sin q, is obtained as b sin θ = a cos θ a tan θ = b a θ = tan −1 b

r = b sin q q = tan−1 a b P

B

r = a cos q

A O

q=0

a Hence, θ = tan −1 at P. b 4. Divide the region into two subregions OAP and OBP. Draw an elementary Fig. 9.93 radius vector OA and OB in each subregion. (i) In subregion OAP, elementary radius vector OA starts from the origin and terminates on the circle r = b sin q. Limits of r : r = 0 to r = b sin q a Limits of q : q = 0 to θ = tan −1 b (ii) In subregion OBP, elementary radius vector OB starts from the origin and terminates on the circle r = a cos q. Limits of r : r = 0 to r = a cos q a π Limits of θ : θ = tan −1 to θ = b 2 Hence, the polar form of the given integral is

I = ÚÚ =Ú

( x 2 + y 2 )2 x 2 y2

tan -1

a b

0

=Ú

tan -1

0

a b

dx d y

Ú0

p

r4

b sin q

r 4 sin sin 2 q cos cos2 q

r2 sin n 2 q cos2 q 2 1

b sin q

0

◊ r dr dq + Ú 2 -1 a Ú tan ta

p 2

0

b

r4

a cosq

r 4 sin sin 2 q ccoos2 q

r2 dq + Ú -1 a tan ta sinn 2 q cos2 q 2 b 1

p

a

◊ r dr dq

a cos cosq

dq 0

=

1 tan -1 b 1 1 1 ◊ b2 sin 2 q dq + Ú 2 -1 a ◊ a 2 cco o s 2 q dq Ú 2 2 2 2 2 0 2 ta tan sin n q cos q sin n q c os q b

=

b2 2

tan -1 tan

Ú0

a b sec sec 2 q

dq +

a2 2

p 2

Útatan-1 a cosec q dq 2

b

9.82

Chapter 9

Multiple Integrals

-

+

a2 - cot q 2

b tan tan 2

a b

tan

b 2

2

b 2

a -0 b

a È 2

p 2 tan -1

˘

a b

a2 2

t

a ˘ b

cot tan

2

b a

ab ab + 2 2 ab

Example 5 Evaluate

∞

∫ ∫

x

e

x y

Solution

y

1. Limits of x : x = 0 to x → ∞ Limits of y : y = 0 to y → ∞ 2. The region of integration is the first quadrant. 3. Putting x = r cos q, y = r sin q, the integral changes to polar form. 4. Draw an elementary radius vector which starts from the origin and extends up to infinity. Limits of r : r = 0 to r → ∞ Limits of q : q = 0 to

=

Fig. 9.94

2 Hence, the polar form of the given integral is ∞

p 2

Ú Ú

x

O

q

p 2

r→

y

• -r

e

r dr dq

p

-

1 2 • -r 2 e 2 Ú0 Ú

-

1 2 -r 2 • e dq 2 Ú0 0

-

1 2 ( 2 Ú0

p

p

r )dr dq ∵Ú

f ¢(r r

ef

r)

O

q=0

e )dq Fig. 9.95

9.5

Change of Variables

9.83

p 2 0

1 = - -q 2 p = 4

Example 6

∫ ∫

Evaluate

∞

(

)

3 2

Solution 1. Limits of x x → −∞

to

x→∞

Limits of y y → −∞

to

y→∞

2. The region of integration is the entire coordinate plane. 3. Putting x = r cos q, y = r sin q, integral changes to polar form. 4. Draw an elementary radius vector which starts from origin and extends up to •. Limits of r : r = 0 to r → ∞ Limits of q : q = 0 to q = 2p Hence, the polar form of the given integral is ∞ ∞ d y I=∫ ∫ 3 ∞ ∞ ( + y )2 2 r d

∫ ∫

0

1 = 2

2

1 2

2

=

Fig. 9.96

3

∞

−

+

3

)dr d

−2 1 +

)

1 2

q= p 2

r→∞

d 0

= −∫

2

(r )]n 1

1+ r2 2

= −∫ ( =2

x

( + r )2

∵∫

=

y

r )dr

f r )]n +1 n +1

O

d 0

)d

0

Fig. 9.97

=0

9.84

Chapter 9

Multiple Integrals

Example 7 Evaluate ∫

a

∫

x

a

x

y

2

y2

dx y by transforming into polar coordinates.

[Winter 2013] Solution y

1. Limits of x : x = y to x = a Limits of y : y = 0 to y = a 2. The region of integration is bounded by the lines y = x, x = a and y = 0. 3. Putting x = r cos q, y = r sin q, polar form of (i) the line y = x is r sin q = r cos q, tan

Q (a a)

x=a

q = 1,

. 4 (ii) the line x = a is r cos q = a, r = a sec q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the line r = a sec q. Limits of r : r = 0 to r = a sec q Limits of q : q = 0 to

P

O

x

Fig. 9.98

=

4 Hence, the polar form of the given integral is x

a

I

y

x c

=

0

∫∫ 4

a se

y

dx y

r cos ⋅r r2

y

d

=

p 2

cos Q

=∫ r0

cos

= ∫ 4 a se

cos d

a se

4

A r = a secq

= a∫ 4 d a

4 0

⋅

4 a 4

P

O

Fig. 9.99

=0

9.5

Change of Variables

9.85

Example 8 Evaluate

x

2 x x2

2

∫∫

x

2

x by transforming into polar coordinates.

y2

Solution 1. Limits of y : y = 0 to x2 Limits of x : x = 0 to x = 2 2. The region of integration is bounded by the circle x 2 + y 2 – 2x = 0 and the lines y = 0, x = 0. Since the limits of x and y are positive, the region of integration is the part of the circle in the first quadrant. 3. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y 2 – 2x = 0 is

y

x 2+y 2 2x = 0

(2, 0) x

O

r2 – 2r cos q = 0 r = 2 cos q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 2 cos q. Limits of r : r = 0 to r = 2 cos q Limits of q : q = 0

Fig. 9.100

to

=

2 Hence, the polar form of the given integral is I

2 x x2

2 0

x

p 0

Ú

dy x

r = 2 cosq

q

r2

(2, 0) O

Ú Ú p 2

y

2

r cos q

2 cosq 0

p 2

A

x 2

p 2

2 co q

2

r0

cos q

q

=0

q

cos q q

p

Ú 2 2 co

2

q q

p

+ cos q q sin 2q + 2

p 2 0

Fig. 9.101

9.86

Chapter 9

Multiple Integrals

p 1 1 + sinn p - sin 0 2 2 2 p = 2 =

Example 9 2 x − x2

1

∫∫

Evaluate

0

x

((xx 2 + y 2 )dx dy. y

Solution y=x

1. Limits of y : y = x to y = 2 x − x 2 Limits of x : x = 0 to x = 1 2. The region of integration is bounded by the line y = x and the circle x2 + y 2 – 2x = 0. O 3. Putting x = r cos q, y = r sin q, polar form of (i) the line y = x is π r sin θ = r cos θ , tan tan θ = 1, θ = . 4 2 2 (ii) the circle x + y – 2x = 0 is r2 – 2r cos q = 0 r = 2 cos q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 2 cos q. Limits of r : r = 0 to r = 2 cos q π π to θ = 4 2 Hence, the polar form of the given integral is

P (1, 1)

x 2 + y 2 − 2x = 0

(1, 0)

x

Fig. 9.102 q= p 2 A

P

q= p 4 r = 2 cos q

Limits of θ : θ =

I=Ú

1

0 Úx

=Ú =Ú

2 x - x2

q=0

( x 2 + y 2 )d dxx dy

p 2 cosq 2 2 r ◊ r dr dq p 0 4 p 4 2 cosq 2 r dq p 4 0 4

Ú

p

= 4Ú p2 cos4q dq 4

O

Fig. 9.103

9.5

=

p 2 Ê 1 + cos 2q ˆ 2 4 p Á ˜¯ dq Ë 2 4 p 2 2q q + cos2 2q ) dq p (1 + 2 cos 2 4 p 1 + cos 4q ˆ 2Ê ˜¯ p Á 1 + 2 cos 2q + Ë 2 4 p 3 2 sin 2q sin sin 4q 2

Change of Variables

9.87

Ú

=Ú =Ú =

2

q+

2

+

8

dq

p 4

=

3Êp pˆ Ê pˆ 1 - sin p - sin ˜ + (sin 2p - sin sin p ) 2 ÁË 2 4 ˜¯ ÁË 2¯ 8

=

3p +1 8

Example 10 Evaluate the integral

a

Ú0 Ú0

a2 - y2

y 2 x 2 + y 2 dy dx by changing into [Summer 2014]

polar coordinates. Solution 1. Limits of x : x = 0 to x =

a2 - y2

Limits of y : y = 0 to y = a 2. The region of integration is bounded by x = 0, x=

a 2 - y 2 , y = 0 and y = a.

3. Putting x = r cos q, y = r sin q, polar form of the circle x2 + y2 = a2 is (i) r2 = a2 \r=a 4. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = a. Limits of r: r = 0 to r = a p Limit of q: q = 0 to q = 2

Fig. 9.104

Fig. 9.105

9.88

Chapter 9

Multiple Integrals

Hence, the polar form of the given integral is I= =

=

p a 2 2 r 0 0

Ú Ú

sin 2 q ◊ r ◊ r dr dq

p 2 0

a sin 2 q ÈÍ Ú r 4 dr ˘˙ dq 0 Î ˚

p 2 0

r5 sin q 5

Ú Ú

a

2

p 2 0

=

a5 5

=

a5 1 p ◊ ◊ 5 2 2

=

a 5p 20

Ú

dq 0

sin 2 q dq

Example 11 2

Evaluate

1

Ú0 Ú

1- x 2 x - x2

2

xye - ( x + y ) dx dy. x 2 + y2

Solution 1. Limits of y : y = x − x 2

to

y = 1 − x2

Limits of x : x = 0 to x = 1 2. The region of integration is the part of the first quadrant bounded by the circles x 2 + y 2 – x = 0 and x 2 + y 2 = 1. 3. Putting x = r cos q, y = r sin q, polar form of (i) the circle x 2 + y 2 – x = 0 is r2 – r cos q = 0, r = cos q. (ii) the circle x2 + y2 = 1 is r2 = 1, r = 1. 4. Draw an elementary radius vector OAB from the origin which enters in the region from the circle r = cos q and terminates on the circle r = 1. Limits of r : r = cos q to r = 1 p Limits of q : q = 0 to q = 2

y

x2 + y2 = 1 x2 + y2 – x = 0 P (1, 0) O

x

Fig. 9.106

9.5

Hence, the polar form of the given integral is I=Ú

1

Ú

p

=

1 x2 x x

xye -

2

◊

r2

p

=-

1 2 n 2Ú

=-

1 2 sin cos q e 2Ú

qÚ

1

p

Úe

r2

=-

1 21 1 sin 2Ú 2 e

O

r ) dr dq

q

cos q

0

dq q

r

p

1 2 sin 2Ú

r

A

1

f ¢(

=-

r dq

r2

e

q

r B

r2

e

co q

p 2

dx y

y2

r2

1

q

y )

x

x2

9.89

Change of Variables

q (e

e

)

ef cos2 q

Fig. 9.107

)dq

p

q cos q dq p

2q 2

1 1 =4 e =-

1 4

1 2e

=-

1 4

1 2e

e

cos

(q )

f 0

p - co 0)

- co

2p

+e

cos2 0

-1 ˘

2

1 1 1 - + 4 e e 1 4

2˘ e

Example 12 Evaluate

4a

y

Ú Úy

2

4a

x2 x2

y2 dx y y2

Solution y2 to x = y 4a Limits of y : y = 0 to y = 4a.

1. Limits of x x

[Winter 2013]

9.90

Chapter 9

Multiple Integrals

2. The region of integration is bounded by the line y = x and the parabola y 2 = 4ax. 3. Putting x = r cos q, y = r sin q, polar form of

y y2

P(4a, 4a)

4ax

(i) the line y = x is r sin q = r cos q, p tan q = 1, q . 4 (ii) the parabola y2 = 4ax is r 2 sin2 q = 4ar cos q, r = 4a cot q cosec q. 4. Draw an elementary radius vector OA which starts from the origin and terminates on the parabola r = 4a cotq cosecq.

x

r = 4a cot q cosec q

Limits of r : r = 0 to Limits of q q =

O

p to 4

p 2

q

Fig. 9.108

Hence, the polar form of the given integral is I

4a

y

x2

2

0

y 4a

x2

y2

=Ú =

p 2 p 4 p

Ú

q

dx y

r = 4a cot q cosec q P

r2

2

A

q - sin

)

r2 - 2 sin 2 q

p 4

1 2

4 a cot q c

p 2 p 4

q= p 2

r2 2

◊r

dq q

p 4

4 a cot q cosec q

dq 0

O

- 2 sin 2 q 4 a ) c

2

q

=0

q dq

p

q

co

p 4

p

=

-

p 4

-

-

+2

q Fig. 9.109

cot 3 q + 3 2

-

3

co

+2

p 2

f( ]

(q )dq =

p 4

p p - cot + 2 4

p p - cot + 2 4

p p 2 4

f (q )]n +1 ˘ n +1 ˚

9.5

Change of Variables

9.91

p˘ È 1 = 8a 2 Í- (-1) + 2(-1) + 2 ◊ ˙ 4˚ Î 3 È 5 p˘ = 8a 2 Í - + ˙ Î 3 2˚ Èp 5 ˘ = 8a 2 Í - ˙ Î 2 3˚

Example 13 Evaluate

a

Ú0 Ú2

5 ax - x 2

x2 + y 2 y2

ax

dx dy. y

Solution 1. Limits of y : y = 2 ax to y = 5a axx − x Limits of x : x = 0 to x = a 2. Since the limits of x and y are positive, the region of integration is the part of the first quadrant bounded by the parabola y 2 = 4ax and the circle x2 + y 2 – 5ax = 0 3. Putting x = r cos q, y = r sin q, polar form of

y y 2 = 4ax

2

(i) the parabola y2 = 4ax is r2 sin2 q = 4ar cos q, r = 4a cot q cosecq. (ii) the circle x2 + y2 – 5ax = 0 is r2 – 5ar cos q = 0, r = 5a cos q. 4. The points of intersection of r = 4a cotq cosecq and r = 5a cos q are obtained as 4a cot q cosec q = 5a cos q 4 sin 2 q = 5 2 q = ± sin -1 5 Hence, q = sin -1

2 5

P

x 2 + y 2 – 5ax = 0

O

x

Fig. 9.110 q= p 2 B

r = 4a cotq cosecq P r = 5a cosq

A

O

q=0

Fig. 9.111

at P.

9.92

Chapter 9

Multiple Integrals

5. Draw an elementary radius vector OAB from the origin which enters in the region from the parabola r = 4a cotq cosecq and terminates on the circle r = 5a cos q. Limits of r : r = 4a cotq cosecq to r = 5a cos q p 2 Limits of q to q = sin -1 2 5 Hence, the polar form of the given integral is a

Ú Ú2

I

dx y r

5 a cosq 2

1

4 a cot q cosecq

5

p 2

Úsin

y2

y2

ax

p 2

Úsin

x2

5 ax - x

q

cosec q

2

1

r r dq

r 2 n2 q

a

q

t

5

p

= n

1

-

2

cosecq )dq

5

p

= n

+

1

-5

a p 2

= -5a

È

cosec q 3

-

p 2 n

1

dq

cosec

q )dq

[ f (q )]n +1 ˘ n+

5

aÊ 5ˆ 3 2

)

È ∵Ú f

2

acosec sin

5 4a + 2 3

= -5 a 3

q

5 3

=

2

c

5 3˘

3

∵ cosec

2

-

1

a 3

3

2 ˆ

sin

cose

5 =

1

2 5

cosec -1

5 ˘ 2

5 2

Example 14 Evaluate

ÚÚ xy

x2 y 2 + =1 a2 b

2

n 2ˆ2

x y + 2 2 a b

over the first quadrant of the ellipse

9.5

Change of Variables

9.93

Solution Let

x = ar cos q, y = br cos q ∂x ∂x ∂( x, y) ∂r ∂q J= = ∂(r , q ) ∂y ∂y ∂r ∂q a cos q - aarr sin q = = abr b sin q bbrr cos q

y

x

O

dx dy = J dr dq = abr dr dq

Under the transformation x = ar cos q, x2 y 2 y = br sin q, the ellipse 2 + 2 = 1 in the a b xy-plane gets transformed to r 2 = 1 or r = 1, circle with centre (0, 0) and radius 1 in the rq -plane. The region of integration is the part of the circle r = 1 in first quadrant in the rq -plane. Draw an elementary radius vector OA which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0 to r = 1 p Limits of q : q = 0 to q = 2 Hence, the polar form of the given integral is

Fig. 9.112 q=

p 2 A r=1

O

n

Ê x 2 y2 ˆ 2 I = ÚÚ xxyy Á 2 + 2 ˜ dx dy b ¯ Ëa p

Fig. 9.113 n

1

= Ú 2 Ú abr 2 cos q sin q (r 2 ) 2 abr dr dq 0

0

p

= a 2 b2 Ú 2 0

sin 2q 1 n + 3 (r ) dr 2 Ú0

a 2 b2 cos 2q = 2 2 2 2

p 2 0

r n+4 n+4

1

0

=

a b 1 (- cos p + cos 0) ◊ 4 n+4

=

a 2 b2 2(n + 4)

q=0

9.94

Chapter 9

Multiple Integrals

EXERCISE 9.5 Change to polar coordinates and evaluate the following integrals: 1.

1

∫∫

dx dy over the region bounded by the semicircle x 2 + y 2 – x = 0,

xy y ≥ 0.

2.

∫∫ y

2

 p   Ans. :  2 

dx dy over the area outside the circle x 2 + y 2 – ax = 0 and inside the

circle x 2 + y 2 – 2ax = 0.

3.

∫∫ sin(x

 15p a 4   Ans. :  64  

+ y 2 )dx dy over the circle x2 + y2 = a2.

2

 Ans. : p((1 − ccos a 2 ) 3

4.

2 2 ∫∫ xy(x + y )2 dx dy over the first quadrant of the circle x2 + y2 = a2.

 a7   Ans. :  14   5.

6.

3

3x

∫∫ 0

0

a

x

0

0

Ú Ú

dy dx x2 + y2

3    Ans. : 2 log 3  

x 3 dx dy x2 + y2

  a4  Ans. : log 1 + 2  4  

(

7.

8.

a

∫ ∫ 0

0

a

∫ ∫ 0

a2 − x 2

0

a2 − x 2

p  sin n  2 (a 2 − x 2 − y 2 ) dx dy a 

e−x

2

−y2

)

 a2   Ans. :  2 

dx dy

(

)

p  −a 2   Ans. : 4 1 − e   

9.5

9.

0

ÚÚ 0

12.

2

0

Ú Ú

a2 - y 2

2

0

y

1˘ È Í Ans. : e ˙ Î ˚

log g e(x 2 + y 2 )dx dy È p 2Ê 1ˆ ˘ Í Ans. : a ÁË log a - ˜¯ ˙ 4 2 Î ˚

a

a + a2 - y 2

0

y

Ú Ú

9.95

È 3p a 4 ˘ Í Ans. : ˙ 4 ˚ Î

2 2 4 xy e -( x + y ) dx dy 2 x +y

x - x2

a

11.

(x 2 + y 2 )dx dy

0

1

10.

2 ax − x 2

2a

∫ ∫

Change of Variables

dx dy (4 a + x 2 + y 2 )3 2

È 1 Êp 1 1 ˆ˘ tan-1 Í Ans. : 2 Á ˜˙ 8a Ë 4 2 2¯˚ Î 13.

a

a2 - x 2

0

ax - x 2

Ú Ú

dx dy a - x2 - y2 2

ÎÈAns. : a ˘˚ a2 - x 2

a

14.

Ú Ú

ax - x

0

2

2 2 xy e -( x + y ) dx dy 2 x +y

2

1 È È 2 - a2 ˘ ˘ Í Ans. : 4 a 2 Î1 - (1 + a )e ˚ ˙ Î ˚ 15.

1

x

dx dy

0

x2

x2 + y2

ÚÚ

È Ans. : 2 - 1˘ Î ˚

9.5.2

Change of Variables from Cartesian to Other Coordinates

In some cases, evaluation of double integral becomes easier by changing the variables. Let the variables x, y be replaced by new variables u, v by the transformation x = f1 (u, v), y = f2 (u, v), then

ÚÚ f ( x, y)dx dy = ÚÚ f ( f1 , f2 ) J dudv where

∂x ∂( x, y) ∂u Jacobian J = = ∂(u, v) ∂y ∂u

∂x ∂v ∂y ∂v

Using Eq. (1), the double integral can be transformed to new variables.

... (1)

9.96

Chapter 9

Multiple Integrals

Example 1 Ê x - yˆ dxdy Using the transformation x - y = u, x + y = v, evaluate ÚÚ cos Á Ë x + y ˜¯ over the region bounded by the lines x = 0, y = 0, x + y = 1. x - y = u, x + y = v u+v v-u x= ,y = 2 2

Solution

∂x ∂( x, y) ∂u J= = ∂(u, v) ∂y ∂u

∂x 1 ∂v 2 = ∂y 1 ∂v 2 1 dxdy = J dudv = dudv 2

1 1 1 1 2 = + = 1 4 4 2 2

The region bounded by the lines x = 0, y = 0 and x + y = 1 in xy-plane is a triangle OPQ. Under the transformation x =

and = , 2 2 (i) the line x = 0 gets transformed to the line u = –v y v Q

v=1

C Q' A

x+y=1

x=0

P' B

u = −v

u=v

P O

x

O

u

y=0

Fig. 9.114

(ii) the line y = 0 gets transformed to the line u = v (iii) the line x + y = 1 gets transformed to the line v = 1 Thus, triangle OPQ in xy-plane gets transformed to triangle OP'Q' in uv-plane bounded by the lines u = v, u = –v and v = 1. In the region, draw a horizontal strip AB parallel to u-axis which starts from the line u = –v and terminates on the line u = v. Limits of u : u = –v to u = v Limits of v : v = 0 to v = 1

9.5

Change of Variables

9.97

Ê x - yˆ I = ÚÚ cos Á dx dy Ë x + y ˜¯ =Ú

1 v

Ú

0 -v

=

Ê uˆ 1 cos Á ˜ du dv Ë v¯ 2

1 1 Ê uˆ v sin Á ˜ Ë v¯ 2 Ú0

v

dv -v

=

1 1 v [sin 1 - sin(-1)] dv 2 Ú0

=

1 v2 ◊ 2 sin 1 2 2

1

0

1 = sin 1 2

Example 2

2 2 Using the transformation x 2 - y 2 = u , 2 xy = v, find ÚÚ ( x + y ) dx dy over the region in the first quadrant bounded by x2 – y2 = 1, x2 – y2 = 2, xy = 4, xy = 2.

Solution x2 – y2 = u, 2xy = v It is difficult to express x and y in terms of u and v, therefore we write Jacobian of u, v in terms of x and y.

∂u ∂ (u, v) ∂x J= = ∂ ( x , y ) ∂v ∂x

∂u ∂y 2x - 2y = = 4 ( x 2 + y2 ) ∂v 2y 2x ∂y

(

)

dudv = J dxdy = 4 x 2 + y 2 dxdy dxdy =

(

1

4 x + y2 2

)

dudv

The region in xy-plane bounded by the curves x2 – y2 = 1, x2 – y2 = 2, xy = 4, xy = 2 is transformed to a square in uv-plane bounded by the lines u = 1, u = 2, v = 4, v = 8. In the region, draw a vertical strip AB parallel to the v-axis which starts from the line v = 4 and terminates on the line v = 8. I = ÚÚ ( x 2 + y 2 ) dx dy

9.98

Chapter 9

Multiple Integrals

=Ú

2 8

Ú4 ( x

1

+ y2 )

2

1 4 ( x + y2 ) 2

du dv

1 2 8 u v 4 1 4 =1 =

v

y xy = 4 xy = 2

v=8 R'

B

Q'

S'

A

P'

x2 − y2 = 1

R Q

x2 − y2 = 2

S P

u=2

u=1 x

O

O

v=4 u

Fig. 9.115

Example 3 Evaluate

Ú Ú (x

2

+ y 2 ) dA by changing the variables, where R is

R

the region lying in the first quadrant and bounded by the hyperbola [Summer 2014] x2 – y2 = 1, x2 – y2 = 9, xy = 2 and xy = 4. Solution Let

u = x2 – y2

J=

v = xy ∂u ∂x

∂(u, v) = ∂v ∂( x, y) ∂x

∂u ∂y ∂v ∂y

=

2 x -2 y y x

= 2x2 + 2y2 = 2(x2 + y2) du dv = |J| dx dy = 2(x2 + y2) dx dy 1 dx dy = ◊ du dv 2 2( x + y 2 ) The region in the xy-plane bounded by the curves x2 – y2 – 1, x2 – y2 = 9, xy = 2 and xy = 4 is transformed to a square in the uv-plane bounded by the lines u = 1, u = 9, v = 2, v = 4.

9.5

Change of Variables

9.99

v

y xy = 4 xy = 2

v=8 x2 − y2

R Q

R'

B

Q'

S'

A

P'

=1

x2 − y2 = 9

S P

u=2

u=1 x

O

O

v=4 u

Fig. 9.116

I=

ÚÚ ( x

2

+ y 2 ) dx dy

ÚÚ ( x

2

+ y2 ) ◊

R

=

R

=

1 2

1 = 2

1 2( x + y 2 ) 2

du dv

ÚÚ du dv 49

Ú Ú du dv 21

1 4 9 v u1 2 2 1 = (4 - 2) (9 - 1) 2 1 = (2)(8) 2 =8 =

Example 4 Using the transformation x + y = u, y = uv, show that 1 1- x

Ú0 Ú0

e

y x+ y

dy dx dx =

1 (e - 1). 2

[Winter 2014]

9.100

Chapter 9

Multiple Integrals

Solution

x + y = u, y = uv x = u(1 - v), y = uv

∂x ∂x ∂( x, y) ∂u ∂v 1 - v - u J= = = = (1 - v)u + uv = u v u ∂(u, v) ∂y ∂y ∂u ∂v dxdy = J du udv = udu dv Limits of y : y = 0 to y = 1 – x Limits of x : x = 0 to x = 1. The region in xy-plane is the triangle OPQ bounded by the lines x = 0, y = 0 and x + y = 1. Under the transformation x = u (1 – v) and y = uv, (i) the line x = 0 gets transformed to the line u = 0 or v = 1 (ii) the line y = 0 gets transformed to the line u = 0 or v = 0 (iii) the line x + y = 1 gets transformed to the line u = 1 v y

v=1 R'

Q

B

Q' (1, 1)

u=0

u

1

P x

O

O

A v

P'

u

0

Fig. 9.117

Thus, the triangle OPQ in the xy-plane gets transformed to the square OP'Q'R' in uv-plane bounded by the lines u = 0, v = 0, u = 1 and v = 1. In the region, draw a vertical strip AB parallel to the v-axis which starts from the u-axis and terminates on the line v = 1. Limits of v : v = 0 to v = 1 Limits of u : u = 0 to u = 1 y I=Ú

1 1- x

=Ú

1 1 v

Ú

0 0

Ú

0 0

= ev

1 0

e x + y dx dy

e udu dv u2 2

1

0

= (e1 - e0 ) ◊ =

1 (ee -1) 2

1 2

9.5

Change of Variables

9.101

Example 5 Evaluate

u=

4

Ú0 Ú

y +1 2 2x y 2 2

y

dx dy by applying the transformations

y 2x - y , v = . Draw both regions. 2 2

[Winter 2015]

Solution 1. The function is integrated first w.r.t. x. y y to x = + 1 2. Limits of x : x = 2 2 Limits of y : y = 0 to y = 4. 3. The region is the parallelogram bounded y y , x = + 1, y = 0 and by the lines x = 2 2 y = 4 in xy-plane. Applying the transformations 2x - y y , v= 2 2 y = x2 y the line x - = 0 mapped to the line u = 0 2 y the line x - = 1 mapped to the line u = 1 2 the line y = 0 mapped to the line v = 0 the line y = 4 mapped to the line v = 2 u=

(i) (ii) (iii) (iv)

Fig. 9.118

Hence, the parallelogram OABC in the xy-plane mapped to the rectangle O¢A¢B¢C¢ in uv-plane, bounded by the lines u = 0, u = 1, v = 0 and v = 2. Fig. 9.119 In the region, draw a vertical strip AB parallel to the v-axis which starts from the u-axis and terminates on the line v = 2. Limits of u : u = 0 to u = 1 Limits of v : v = 0 to v = 2 dxdy = J du dv where Jacobian, J = J* =

∂( x, y) ∂(u, v) 1 ∂(u, v) = J ∂( x, y)

9.102

Chapter 9

Multiple Integrals

∂u ∂x = ∂v ∂x

∂u ∂y ∂v ∂y

1 = 0

1 2 1 2

1 2 1 \ dxdy = dudv 2 Hence, the new form of the integral is 4 y +1 2 x - y I = Ú Úy dxdy 0 2 2 =

=Ú

2

Ú

1

v=0 u=0

1 u ◊ du dv 2 1

1 2 u2 = Ú dv 2 0 2 0

1 2 = v0 4 1 = (2 ) 4 =2

Example 6 Using the transformation x = u (1 + v), y = v(1 + u ), u ≥ 0, v ≥ 0, evaluate Solution

2

y

Ú0 Ú0 ÈÎ( x - y)

2

-

1 2

+ 2( x + y ) + 1˘˚ dydx. x = u(1 + v), y = v(1 + u)

∂x ∂x ∂( x, y) ∂u ∂v 1 + v u J= = = = 1+ u + v v 1+ u ∂(u, v) ∂y ∂y ∂u ∂v dxdy = J dudv = (1 + u + v) dudv

9.5

Change of Variables

9.103

v y

y=x

y=2

v (1 + u) = 2

Q P

Q'

x=0

P'(1, 1) A

x

O

v=u

B

u = −1 u=0

u

O

Fig. 9.120

Limits of x : x = 0 to x = y Limits of y : y = 0 to y = 2. The region in the xy-plane is the DOPQ bounded by the lines x = 0, y = 2 and y = x. Under the transformation x = u (1 + v), y = v (1 + u), u ≥ 0, v ≥ 0 (i) the line x = 0 gets transformed to the line u = 0 (ii) the line y = 2 gets transformed to the curve v (1 + u) = 2 (iii) the line y = x gets transformed to the line u = v Thus, the triangle OPQ in the xy-plane gets transformed to the region OP'Q' in uv plane bounded by the lines u = 0, u = v and the curve v (1 + u) = 2. The point of intersection of u = v and v (1 + u) = 2 is obtained as u2 + u – 2 = 0, u = 1, –2 and v = 1, –2. The point of intersection is P' (1,1). In the region, draw a vertical strip AB parallel to the v-axis which starts from the line u = v and terminates on the curve v (1 + u) = 2. 2 Limits of v : v = u to v = 1+ u Limits of u : u = 0 to u = 1 I=Ú

2

Ú

y

0 0

È( x - y) + 2( x + y) + 1˘ Î ˚

2 1+ u 0 u

=Ú

1

Ú

2

-

1 2

dy dx -

È(u - v)2 + 2(u + v + 2uv) + 1˘ Î ˚

2 1+ u (1 + u + v ) -1 (1 + u + v ) dv du 0 u

=Ú

1

Ú

2 1+ u dv du u 0 u

=Ú

1

Ú

1

2 1+ u

0

u

=Ú v

du

1Ê 2 ˆ =Ú Á - u˜ du 0 Ë 1+ u ¯

1 2

(1 + u + v)du dv

9.104

Chapter 9

Multiple Integrals

u2 = 2 log(1 + u) 2 = 2 log 2 -

1

0

1 2

Example 7 Evaluate ÚÚ xy dxdy by changing the variables over the region in the first quadrant bounded by the hyperbolas x2 – y2 = a2, x2 – y2 = b2 and the circles x2 + y2 = c2, x2 + y2 = d2 with 0 < a < b < c < d. Solution Let

x 2 - y 2 = u, x 2 + y 2 = v u+v 2 v-u x2 = ,y = 2 2 1 1 ∂x ∂x 1 ∂( x, y) ∂u ∂v 4x 4x = = J= = 1 1 8 xy ∂(u, v) ∂y ∂y 4y 4y ∂u ∂v dxdy = J dudv = xydxdy =

1 dudv 8 xy

dudv 8

The region bounded by the hyperbolas x2 – y2 = a2, x2 – y2 = b2 and the circles x2 + y2 = c2, x2 + y2 = d 2 in xy-plane is the curvilinear rectangle PQRS. v y S'

B

v = d2

R'

x2 − y2 = a 2 x2 − y2 = b 2

R Q

S

u P

x2 + y 2 = d2 x x2 + y 2 = c2

Fig. 9.121

u = b2

a2 P' O

A

v = c2

Q' u

9.5

9.105

Change of Variables

Under the transformation x2 – y2 = u and x2 + y2 = v, (i) the hyperbolas x2 – y2 = a2, x2 – y2 = b2 get transformed to the lines u = a2, u = b2 respectively. (ii) the circles x2 + y2 = c2, x2 + y2 = d 2 get transformed to the lines v = c2, v = d 2 respectively. Thus, the curvilinear rectangle PQRS in the xy-plane gets transformed to the rectangle P'Q'R'S' in uv-plane bounded by the lines u = a2, u = b2, v = c2 and v = d 2. In the region, draw a vertical strip AB parallel to v-axis which starts from the line v = c2 and terminates on the line v = d 2. Limits of v : v = c2 to v = d 2 Limits of u : u = a2 to u = b2 I = ÚÚ xydxdy =Ú

b2

u=a

d2

2

Úv = c

2

1 dudv 8

1 b2 d 2 = u a2 v c2 8 1 = (b 2 - a 2 )(d 2 - c 2 ) 8

Example 8 2 Evaluate ∫∫ ( x + y ) dx dy, by changing the variables parallelogram with vertices (1, 0), (3, 1), (2, 2), (0, 1).

over

the

Solution The region of integration in xy-plane is the parallelogram PQRS. Equations of the sides of the parallelogram are obtained as (i) PQ : y - 0 =

1- 0 ( x -1) 3 -1

y

2 y = x -1 x - 2y =1 2 -1 ( x - 0) 2 -0 2y -2 = x x - 2 y = -2

(ii) RS : y -1 =

(iii) PS : y - 0 = 1 - 0 ( x -1) 0 -1 x + y =1

R (2, 2) S (0, 1)

O

Q (3, 1)

P (1, 0)

Fig. 9.122

x

9.106

Chapter 9

Multiple Integrals

2 -1 ( x - 3) 2 -3 y -1 = - x + 3 x+y =4

(iv) QR : y -1 =

Let x - 2 y = u, x + y = v x=

u + 2v v -u ,y= 3 3

∂x ∂( x, y) ∂u = J= ∂(u, v) ∂y ∂u

∂x 1 2 ∂v 3 3 1 = = ∂y 1 1 3 ∂v 3 3 1 dxdy = J dudv = du udv 3

Under the transformation x – 2y = u, and x + y = v (i) the lines x – 2y = 1, x – 2y = –2 get transformed to the lines u = 1, u = –2 respectively. (ii) the lines x + y = 1, x + y = 4 get transformed to the lines v = 1, v = 4 respectively Thus, the parallelogram PQRS in the xyv v=4 plane gets transformed to a square P'Q'R'S' in uv-plane bounded by the lines u = 1, u = R' B Q' –2, v = 1 and v = 4. In the region, draw a vertical strip AB parallel to v-axis which starts from the line v = 1 and terminates on the line v = 4. Limits of v : v = 1 to v = 4 u=1 u = −2 Limits of u : u = –2 to u = 1 I = ÚÚ ( x + y)2 dxdy =Ú

1

Ú

4

u = 2 v =1

v2

1 1 v3 = u -2 3 3

1 dudv 3

A

S'

P'

v=1

4

u

O 1

Fig. 9.123

= 21

Example 9 n

Ê x2 y 2 ˆ 2 Evaluate ÚÚ xy Á 2 + 2 ˜ over the first quadrant of the ellipse b ¯ Ëa x2 y 2 + = 1. a 2 b2

9.5

Change of Variables

9.107

Solution Let x = ar cos q, y = br sin q

y

∂x ∂x ∂( x, y) ∂r ∂q J= = ∂(r , q ) ∂y ∂y ∂r ∂q a cos q - ar sin q = = abr b sin q br cos q

x

O

dx dy = J dr dq = abr dr dq Under the transformation x = ar cos q, x2

+

y2

Fig. 9.124 = 1. in the a 2 b2 xy-plane gets transformed to r2 = 1 or r = 1, circle with centre (0, 0) and radius 1 in the rq -plane. The region of integration is the part of the circle r = 1 in first quadrant in the rq-plane. In the region, draw an elementary radius vector OA from the pole which terminates on the circle r = 1. Limits of r : r = 0 to r = 1

y = br sin q, the ellipse

Limits of q : q = 0 to q =

p 2

q=p 2

n

A

Ê x 2 y2 ˆ 2 I = ÚÚ xy Á 2 + 2 ˜ dxdy b ¯ Ëa p 1 2 abr 2 0 0

=Ú

Ú

p

= a 2 b2 Ú 2 0

n 2 2 cos q sin q (r )

sin 2q 2

a 2 b2 cos 2q = 2 2 2 2

r=1

1

Ú0 (r ) p 2 0

n +3

r n+4 n+4

O

abr dr dq

dr 1

0

=

a b 1 (- cos p + cos 0) ◊ 4 n+4

=

a 2 b2 2(n + 4)

Fig. 9.125

q=0

9.108

Chapter 9

Multiple Integrals

EXERCISE 9.6 1. Using the transformation x + y = u, x – y = v, evaluate

ÚÚ e

x -y x +y

dxdy over

the region bounded by x = 0, y = 0 and x + y = 1. È 1Ê 1ˆ ˘ Í Ans.: ÁË e - ˜¯ ˙ 4 e Î ˚ 2 2 2. Using the transformation x2 – y2 = u, 2xy = v, evaluate ÚÚ (x - y )dxdy over the region bounded by the hyperbolas x2 – y2 = 1, x2 – y2 = 9, xy = 2 and xy = 4.

[Ans.: 4] 3. Using the transformation x + y = u, y = uv, evaluate •

•

0

0

Ú Ú

e -( x + y ) x p -1y q -1dx dy . È Ans.: p q ˘ Î ˚

4. Using the transformation x = u, y = uv, evaluate

1

∫∫

x

0 0

x 2 + y 2 dx dy .

È ˘˘ 1È 2 1 + log 1 + 2 ˙ ˙ Í Ans.: Í 3 ÍÎ 2 2 ÍÎ ˙˚ ˙˚

(

5. Evaluate

)

∫∫ (x + y) dx dy by changing the variables over the region bounded 2

by the parallelogram with sides x + y = 0, x + y = 2, 3x – 2y = 0 and 3x – 2y = 3. 8   Ans. : 5   

6. Evaluate ∫∫ (x − y )4 e x + y dx dy , by changing the variables over the region bounded by the square with vertices at (1, 0), (2, 1), (1, 2), (0, 1).  e3 − e   Ans. :  5   1

7. Evaluate ∫∫ [ xy (1 − x − y )]2 dx dy , by changing the variables over the region bounded by the triangle with sides x = 0, y = 0, x + y = 1. 2p 2 p˘ È Í Ans.: 105 ˙ Î ˚

9.6

Triple Integrals

9.109

9.6 TRIPLE INTEGRALS Let f (x, y, z) be a continuous function defined in a closed and bounded region V in 3-dimensional space. Divide the region V into small elementary parallelopipeds by drawing planes parallel to the coordinate planes. Let the total number of complete parallelopipeds which lie inside the region V be n. Let d Vr be the volume of the rth parallelopiped and (xr, yr, zr) be any point in this parallelopiped. Consider the sum n

S = Â f ( xr , yr , zr )d Vr

…(1)

r =1

d Vr = d xr ◊ d yr ◊ d zr

where,

If we increase the number of elementary parallelopipeds, n, then the volume of each parallelopiped decreases. Hence as n Æ •, d Vr Æ 0. The limit of the sum given by Eq. (1), if it exists is called the triple integral of f (x, y, z) over the region V and is denoted by ÚÚÚÚ f ( x, y, z )dV V

Hence,

ÚÚÚÚ f ( x, y, z)dV = V

where

n

 f ( xr , yr , zr )d Vr n ƕ lim

d Vr Æ 0 r =1

dV = dx dy dz

9.6.1 Triple Integrals in Cartesian Coordinates Triple integral of a continuous function f (x, y, z) over a region V can be evaluated by three successive integrations. Let the region V be bounded below by a surface z = z1 (x, y) and above by a surface z = z2 (x, y). Let the projection of region V in xy-plane be R which be bounded by the curves y = y1(x), y = y2(x) and x = a, x = b. Then the triple integral is defined as b È y2 ( x ) I = Ú ÍÚ a Î y1 ( x )

{Ú

z2 ( x , y ) z1 ( x , y )

}

˘ f ( x, y, z ) dz dy ˙ dx ˚

Note: The order of variables in dx dy dz indicates the order of integration. In some cases this order is not maintained. Therefore, it is advisable to identify the order of integration with the help of the limits.

9.6.2 Triple Integrals in Cylindrical Coordinates Cylindrical coordinates r, q, z are used to evaluate the integral in the regions which are bounded by cylinders along z-axis, planes through z-axis, planes perpendicular to the z-axis.

9.110

Chapter 9

Multiple Integrals

Relations between Cartesian (rectangular) coordinates (x, y, z) and cylindrical coordinates (r, q, f ) are given as x = r cos q y = r sin q z=z z

Then

ÚÚÚÚ

f ( x, y, z )dx dy dz = ÚÚÚ f (r cos q , r sin q , z ) J dz dr dq P (r,, q,, z)

∂x ∂z ∂y ∂z ∂z ∂z

z=z y = r sin q

N

os

q

O rc

y

90°

q

=

r

x

where,

∂x ∂x ∂r ∂q ∂( x, y, y, z ) ∂y ∂y J= = ∂(r , q , z ) ∂r ∂q ∂z ∂z ∂r ∂q cosssq q -r sin q 0 = sin nq n q r cos cos q 0 0 0 1

90° M

Q

x

Fig. 9.126

= cos q (r co coss q ) + r sin n q (si (sin nq ) =r

Hence,

ÚÚÚ f ( x, y, z)dx dy dz = ÚÚÚ f (r cos q , r sin q , z)r dz dr dq

9.6.3 Triple Integrals in Spherical Coordinates Spherical coordinates (r, q, f ) are used to evaluate the integral in the regions which are bounded by the sphere with centre at the origin.

z

P (x,, y,, z)

Relations between cartesian (rectangular) coordinates (x, y, z) and spherical coordinates (r, q, f ) are given as x = r sin q cos f

r q

z = r cos q N

y = r sin q sin f

O f

y

in

q

co

s

y = r sin q sin f = x

Then

rs

q

90°

M

ÚÚÚÚ = ÚÚÚÚ f (r sin q cos cos f , r sin q sin f , r cos q ) J dr dq df f ( x, y, z )dx dy dz

90° in

rs

z = r cos q

f

Q

x

Fig. 9.127

9.6

where

J=

∂x ∂r

∂x ∂q

∂x ∂f

∂( x, y, z ) ∂y = ∂(r ( , q , f ) ∂r

∂y ∂q

∂y ∂f

∂z ∂r

∂z ∂q

∂z ∂f

sinn q cos cos f r cos cossq q cos cos f = sinn q sin sin f cos q

r cos cossq q sin sin f -r sin q

Triple Integrals

9.111

-r sinn q sin sin f r sinn q cos cos f 0

= sin q cos f (r 2 sin sin 2 q cos f ) - r coss q cos cos f (-r -r sin q cos cos f cos q ) - r ssiin q sin sin f ((-rr sin sin 2 q sin sin f - r cos cos2 q sin sin f ) = r 2 sin sinnq q cos cos2 f (sin 2 q + cos cos2 q ) + r 2 sin q sin 2 f = r 2 sin q

Hence,

sin q sin sin f , r cos q )r ÚÚÚÚ f ( x, y, z)dx dy ddzz = ÚÚÚÚ ÚÚ f (r sin q cos f, r sin

2

sin q dr dq df.

Note: If the region of integration is a sphere x2 + y2 + z2 = a2 with centre at (0, 0, 0) and radius a, then limits of r, q, f are (i) For positive octant of the sphere, r : r = 0 to r = a p q : q = 0 to q = 2 p f : f = 0 to f = 2 (ii) For hemisphere, r : r = 0 to r = a

p 2 f : f = 0 to f = 2p

q : q = 0 to q =

(iii) For complete sphere, r : r = 0 to r = a q : q = 0 to q = p f : f = 0 to f = 2p

9.6.4

Change of Variables

In some cases, evaluation of a triple integral becomes easier by changing the variables. Let the variables x, y, z be replaced by new variables u, v, w by the transformation x = f1 (u, v, w), y = f2 (u, v, w), z = f3 (u, v, w).

9.112

Chapter 9

Multiple Integrals

ÚÚÚ f ( x, y, z)dx dy ddzz = ÚÚÚ f ( f1 , f2 , f3 ) J du dv dw

Then

∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u

where,

∂x ∂v ∂y ∂v ∂z ∂v

∂x ∂w ∂y ∂w ∂z ∂w

9.6.5 Working Rule for Evaluation of Triple Integrals 1. 2. 3. 4.

Draw all the planes and surfaces and identify the region of integration. Draw an elementary volume parallel to z ( y or x) axis. Find the variation of z ( y or x) along the elementary volume. Lower and upper limits of z ( y or x) are obtained from the equation of the surface (or plane) where elementary volume starts and terminates respectively. 5. Find the projection of the region on xy (zx or yz) plane. 6. Draw the region of projection in xy (zx or yz) plane. 7. Follow the steps of double integration to find the limits of x and y (z and x or y and z).

Note: (1) If the region is bounded by the cylinders along the z-axis, planes through z-axis, the planes perpendicular to the z-axis, then the cylindrical coordinates are used. (2) If the region is bounded by the sphere, then the spherical coordinates are used.

Type I Evaluation of Triple Integrals when Limits are Given

Example 1 Evaluate

1 2 e

Ú0 Ú0 Ú0 dy dx dz.

Solution 1 2

1 2È e

e

˘

Ú0 Ú0 Ú0 dy dx dz = Ú0 Ú0 ÍÎÚ0 dy ˙˚dx dz =Ú

1 2

Ú

0 0

e

y dx dz 0

1È 2

= Ú Í Ú e dx ˘˙ dz 0Î 0 ˚ 1

2

0

0

= e Ú x dz 1

= e Ú 2 dz 0

9.6

Triple Integrals

9.113

1

= 2e z 0 = 2e

Another method: Since all the limits are constant and integrand (function) is explicit in x, y and z, the integral can be written as 1 2

1

e

2

e

Ú0 Ú0 Ú0 dy dx ddzz = Ú0 dz ◊Ú0 ddxx ◊ Ú0 dy 1

2

e

= z0◊ x0◊ y0 = 1◊ 2 ◊ e = 2e.

Example 2 Evaluate

2 3 2

Ú0 Ú1 Ú1

xy 2 z dz dy dx.

Solution Since all the limits are constant and integrand (function) is explicit in x, y and z, the integral can be written as 2

3 2

Ú0 Ú1 Ú1 xy

2

2

3

2

0

1

1

z dz dy dx = Ú x dx ◊Ú y 2 dy ◊ Ú z dz =

x2 2

= 2◊

2

◊ 0

3

2

1

1

y3 z2 ◊ 3 2

26 3 ◊ 3 2

= 26

Example 3 Evaluate

1 p

p

Ú0 Ú0 Ú0

[Winter 2013]

y sin z dx dy dz.

Solution Since all the limits are constant and integrand (function) is explicit in x, y, and z, the integral can be written as 1 p

p

1

p

p

Ú0 Ú0 Ú0 y sin z dx dy dz = Ú0 sin z dzÚ0 y ddyyÚ0 dx =

1 - cos z 0

y2 ◊ 2

p p

◊ x0 0

9.114

Chapter 9

Multiple Integrals

Ê p2 ˆ = ( - cos 1 + cos 0) Á ˜ (p ) Ë 2 ¯ =

p3 (1 - cos 1) 2

Example 4 1

z 2p

Ú0 Ú0 Ú0

Evaluate

(r 2 cos2 q + z 2 )rdq drdz.

[Winter 2016]

Solution 1

z

2p

Ú0 Ú0 Ú0

(r 2 cos2 q + z 2 )rdq drdz = Ú

1

=Ú

1

z

Ú

0 0

{Ú

2p

0

z

ÏÔ

2p

Ì Ú0 0 Ú0 Ô

=Ú

1

Ó

z

Ú

0 0

= 2p Ú

1

= 2p Ú

1

= 2p Ú

1

0

È 2 Ê 1 + cos 2q ˆ ˘ ¸Ô + z 2 ˙ dq ˝ r drdz Ír ÁË ˜ ¯ 2 Î ˚ ˛Ô

2p ¸ ÏÔ È 1 Ê sin 2q ˆ Ô 2 2 ˘ r r q + z q + Ì Í ˙ ˝ d r dz ÁË ˜¯ 2 2 ˚ 0 Ô˛ ÔÓ Î

0 Ú0

0

}

[r 2 cos2 q + z 2 ] dq r drdz

z

È1 3 2 ˘ Í 2 r + z r ˙ drdz Î ˚

r4 r2 + z2 8 2 z2 z3 + dz 8 2

È z3 z 4 ˘ = 2p Í + ˙ ÍÎ 24 8 ˙˚ È 1 1˘ = 2p Í + ˙ Î 24 8 ˚ È 3 + 1˘ = 2p Í ˙ Î 24 ˚

1

0

2

dz 0

9.6

= 2p ◊ =

Triple Integrals

4 24

p 3

Example 5 Evaluate

1 1 x

Ú Ú

Ú

x y

d

y dz

Solution

ÚÚ Ú

d

z˘d

x+ y

yd 1 1- x

Ú Ú0

x y

z

y)d

0

y2 + 2

1

1

dx

0

1 1 x

1

x

1- x

d 0

+ 2

x

+

(

(

x )2 ˘ dx 2 x )2 ˘ dx 2

3 x2 1 (1 x )3 + ◊ 2 3 2 ( 3)

1

0

1 1 1 - + 2 3 6 1 3

Example 6 Evaluate

1 1- x

Ú Ú

x y

Ú0

e x dy

Solution 1

x

ÚÚ Ú

x

e x y

x

e z˘ d

x

9.115

9.116

Chapter 9

Multiple Integrals

=Ú

1 1- x

=

1

Ú

x y

e

dx

0

1- x

- e )d 1 x

1

0 1

x

dx

1

0

+

x2 - ex 2

1 2

+e

1

0

0

1 2

=

Example 7 Evaluate

a

x

Ú Ú0 Ú

x y x

+

e

dy x

Solution

Ú

a

x

ex

z

a

y x

0

x y

x y

Ú

x y

e

e ez

0

a x

È

a x

=

-

x

x

Î

1 8 1 = 8

◊

e 0

˘

e 2

-

4

e

a

˚

2 + e x dx

2

4

x

x

1 e e = ◊ - ◊ + ex 2 4 2 2 =

x

2x

1 2

a

y

e2 y ◊ 2

x

È x

a

x

x

x

)d - e )d

x

e z˘ d

0

-

4 +

a

-

3 8

)+

-

)

9.6

Triple Integrals

9.117

Example 8 Evaluate the integral

2 2 yz

Ú0 Ú1 Ú0

xyz dx dy dz.

Solution 2

2

yz

Ú0 Ú1 Ú0

xyz dx dy dz = =

=

2

2

2

2

2

2

È

x dx ˘˙ dy dz ˚

yz

Ú0 Ú1 yz ÍÎÚ0 Ú0 Ú1

1

Ú0 Ú1 2 y

1 y4 = 2 4

yz

x2 2

yz

dy dz 0

3 3

z dy dz

2

z4 4

1

2

0

=

1 Ê 1ˆ 1 [16 - 1] [16 - 0] Á ˜ 2 Ë 4¯ 4

=

15 2

Example 9 1 xy x 1 1 0 3

Ú Ú Ú

Evaluate

xy dz dy dx.

Solution 1 xy x 1 1 0 3

Ú Ú Ú

1 x 1 1

xy dz dy ddxx = Ú

3

1 x 1 1

=Ú

3

È

Ú ÍÎÚ0 Ú

xy

˘ dz ˙ xxyy dy dx ˚

xy

z 0 dy dy dx

1 ˘ 3È = Ú Í Ú x xy dy ˙ dx 1 Î 1 ˚

=Ú

3

1

=

x

1 3 x 2y 2

3

dx

1

Ê ˆ 2 3 Á 1 - 1˜ dx x Á 3 ˜ 3 Ú1 Ë x2 ¯

[Summer 2014]

9.118

Chapter 9

Multiple Integrals

=

=

1ˆ 2 3Ê 1 2 dx x Á ˜¯ 3 Ú1 Ë x 3 2x 2

2 log x 3 3

3

1

(

)

3 ˘ 2È 2 Í( log 3 - log 1) - 3 2 - 1 ˙ 3Î 3 ˚ 2È 2˘ = Ílog 3 - 2 3 + ˙ 3Î 3˚

=

Example 10 Evaluate

2 z

yz

Ú0 Ú1 Ú0 xyz dx dy dz.

[Summer 2017]

Solution The innermost limits depend on y and z. Hence, integrating first w.r.t. x, 2

z

yz

Ú0 Ú1 Ú0

xyz dx dy ddzz = Ú

2

Ú

z

0 1

x2 2

yz

yz dy dz 0

1 2 z 2 2 ((yy z ) yz dy dz 2 Ú0 Ú1 z 1 2 = Ú z 3 ÈÍ Ú y3 dy ˘˙ dz ˚ 2 0 Î1 =

z

=

1 2 3 y4 z dzz 2 Ú0 4 1

1 2 = Ú z 3 ( z 4 - 1)d )dz 8 0 =

1 z8 z 4 8 8 4

1 (32 - 4) 8 7 = 2 =

2

0

9.6

Triple Integrals

9.119

Example 11 Evaluate

1 1- x 1- x - y

Ú0 Ú0 Ú0

1 dx dy dz. ((xx + y + z + 1)3

Solution The innermost limits depend on x and y. Hence, integrating first w.r.t. z, 1 1- x 1- x - y

Ú0 Ú0 Ú0

1 ((xx + y + z + 1)3

=Ú

1 1- x È 1- x - y

Ú

0 0

=Ú

dx dy dz

Í Ú0 Î

˘ dz ˙ dy dx ((xx + y + z + +1 1) ˚ 1

3

1 1- x

Ú

0 0

1- x - y

1 -2 2(((xx + y + z + 1)2 2

dy dx 0

=-

˘ 1 1 1- x È 1 1 Í ˙ dy dx Ú Ú 2 2 0 0 2 1- x - y) + 1} ( x + y + 1) ˙˚ ÍÎ {x + y + (1

=-

¸Ô ˘ 1 1 È 1- x ÏÔ 1 1 ÍÚ Ì ˝ dyy˙˙ dx Ú 2 0 ÍÎ 0 ÓÔ 4 ( x + y + 1)2 ˛Ô ˙˚ 1- x

1 1 y 1 =- Ú + dx 0 2 4 x + y +1 0 1 1 È1 - x 1 1 ˘ + dx Í Ú 0 2 Î 4 x + (1 - x ) + 1 x + 1 ˙˚ 1 1Ê 1 - x 1 1 ˆ =- Ú Á + ˜ dx 2 0Ë 4 2 x + 1¯ =-

1 x x2 x =+ - log( x + 1) 2 4 8 2

1

0

1Ê5 ˆ = - Á - log 2˜ ¯ 2Ë8

Example 12 Evaluate

1

Ú0 Ú0

1- x 2

Ú0

1- x 2 - y2

[Winter 2014]

xyz dz dy dx.

Solution 1

Ú0 Ú0

1- x 2

È 1- x2 - y2 ˘ xy Í Ú z dz ˙ dx dy = 0 Î ˚

1

Ú0 Ú0

1- x 2

z2 xy z

1- x 2 - y2

dx dy 0

9.120

Chapter 9

Multiple Integrals

=

1 1 1- x2 xy(1 - x 2 - y 2 ) dx dy 2 Ú0 Ú0

=

˘ 1 1 È 1- x2 È (1 - x 2 )y - y3 ˘˚ ˙ dy dx x ÍÚ Ú Î ˚ 2 0 Î 0

1 1 y2 y 4 = Ú x (1 - x 2 ) 2 0 2 4

1- x 2

dx 0

=

2 Ê 1 - x 2 )2 ˆ ˘ 1 1 È 2 (1 - x ) x (1 x ) Í ÁË ˜ ˙ dx 2 Ú0 Î 2 4 ¯˚

=

1 1 È (1 - x 2 )2 ˘ xÍ ˙ dx 2 Ú0 Î 4 ˚

=

1 1 x(1 - x 2 )2 dx 8 Ú0

=

1 1 {x(1 - 2 x 2 + x 4 )} dx 8 Ú0

=

1 1 ( x - 2 x 3 + x 5 ) dx 8 Ú0

1 x2 2 x 4 x6 + = 8 2 4 6 =

1 Ê 1 1 1ˆ Á - + ˜ 8 Ë 2 2 6¯

=

1 48

1

0

Example 13 Evaluate

e log y e x

Ú1 Ú1

Ú1 logg z dx dy dz.

[Summer 2016]

Solution The inner most limit depends on x and middle limit depends on y. Hence, integrating first w.r.t. z, e

log log y

ex

Ú1 Ú1 Ú1

log z dx dy dz = Ú log

e

Ú

llog og y

1 1

ex

Ú1

llog og z dz dx dy

9.6

e

log y

e È log y

x

1 e

1 dz z ex

+x

gy 1

dy

-2

lo

y

dy

- ex -

1 e

9.121

x dy

z1 d

-

1

e

1

- log 1

x

1 1

1

-Ú z

1

1 1 e

x

x

log y

Triple Integrals

1 dy

e

- 2 ) + g y e - ] dy

1 e

] dy

e

1

2

+

ˆ

1Ê y

1

e

log e

2

1

+e

2

+

ˆ

1

+ e - 1) y 1

e

2

+1

e

-

+y

2

1]

-1

1

e2 2 e2 4

1 4 2e +

+1

13 4

Example 14 Evaluate

• • •

Ú Ú Ú

(

d x

y dz y z

2

Solution

z

1. It is difficult to integrate this integral in cartesian form. Putting x = r sin q cos f, y = r sinq sin f, z = r cosq integral changes to spherical form. xƕ 2. Limits of x x = Limits of y y = Limits of z z

r →∞

O

Æ•

y

•

The region of integration is the positive octant of the plane. Limits of r r r •

x

Fig. 9.128

9.122

Chapter 9

Multiple Integrals

Limits of q = 0

to q =

p 2

p 2 Hence, the spherical form of the given integral is Limits of f : f = 0 to f =

I=Ú

•

=Ú

•

=Ú

•

0

0

0

•

•

Ú0 Ú0

dxdydz ((1 + x + y 2 + z 2 )2

p p 2 2 0 0

Ú Ú

2

r 2 sin q dr dq df ((1 + r 2 )2

r 2 dr (1 + r 2 )2

p 2 0

Ú

p

sin q dq Ú 2 df 0

Putting r = tan t , dr = sec 2 t dt When r = 0, t = 0 When r Æ •, t =

p 2 p

p

tan 2 t

0

0

sec 4 t

I = Ú 2 sin q dq Ú 2 = - cos q

p 2 0

p

sec 2 t dt Ú 2 df

Ê p 2 ˆ ÁË Ú 2 sin t dt ˜¯ f 0

0

p 2 0

ˆ p p Ê ˆ Ê p = Á - coss + cos 0˜ ◊ Á 2 1 - cos 2t dt ˜ ◊ Ú Ë ¯ 2 Ë 0 ¯ 2 2 p

1 sin 2t 2 p = t2 2 0 2 1 Èp 1 ˘p = Í - (sin p - sin 0 )˙ 2 Î2 2 ˚2 2 p = 8

Example 15 Evaluate

a

Ú0 Ú0

a2 - x2

Ú0

a2 - x 2 - y2

xyz dx dy dz.

Solution 1. It is difficult to integrate this integral in cartesian form. Putting sin q sin sin f f,, z = r ccos os q integral changes to spherical form. x = r sin q coss f , y = r sin

9.6

Triple Integrals

2 2 2 2. Limits of z : z = 0 to z = a - x - y

Limits of y : y = 0 to y = a - x Limits of x : x = 0 to x = a 2

z

2

The region of integration is the positive octant of the sphere x 2 + y 2 + z 2 = a 2 .

r

I=Ú

0

Ú0

p

Ú0

a2 - x 2 - y2

p

=Ú2 0

Ú Ú

x

Fig. 9.129

cos q ◊ cos f sin f ◊ r 2 sin q dr dq df

p

a sin 2f 2f df Ú 2 sin sin 3q ccos os q dq Ú r 5 dr 0 0 2

1 - coss 2f = 2 2 =

φ

xyz dx dy dz

p a 2 r 3 sin 2 q f =0 q =0 r =0

=Ú2

y

O

p Limits of q : q = 0 to q = 2 p Limits of f : f = 0 to f = 2 Hence, the spherical form of the given integral is a2 - x 2

P

q

Limits of r : r = 0 to r = a

a

9.123

p 2 0

sin 4 q ◊ 4

p 2

r6 6

0

a

0

È [ f (q )]n +1 ˘˙ Í∵ Ú [ f (q )]n f ¢(q )dq = n +1 ˙ Í Î ˚

1 1 1Ê p ˆ a6 ◊ (- cos p + co oss 0) ◊ Á ssin in - sin 0˜ ◊ ¯ 6 2 2 4Ë 2

1 2 1 a6 ◊ ◊ ◊ 2 2 4 6 a6 = 48 =

Example 16 Evaluate

1

Ú0 Ú0

1- x 2

Ú

dz dy dx dx

1 2

x +y

2

x 2 + y2 + z2

polar coordinates. Solution 2 2 1. Limits of z : z = x + y to

Limits of y : y = 0 Limits of x : x = 0

to to

z=1 y = 1 − x2 x=1

by transforming into spherical

9.124

Chapter 9

Multiple Integrals

2. The region of integration is the part of the cone z2 = x2 + y 2 bounded above by the plane z = 1 in the positive octant (since all three limits are positive). 3. Putting x = r sinq cos f, y = r sinq sin f, z = r cosq, spherical polar form of (i) the cone z2 = x2 + y 2 is r2 cos2q = r2 sin2q (cos2f + sin2f ) = r2 sin2q cosq = sin q tanq = 1 p q= 4 (ii) the plane z = 1 is r cos q = 1 r = secq z

r = sec q A y P

q=

f O

O x

p 4

f

y P

x

Fig. 9.130

4. Draw an elementary radius vector OA which starts from the origin and terminates on the plane r = secq. Limits of r : r = 0 to r = secq p Limits of q : q = 0 to q = 4 p (in positive octant) 2 Hence, the spherical form of the given integral is

Limits of f : f = 0 to f =

I=Ú

1

0 Ú0

1- x 2

p

Ú

p

=Ú2

dz dy dx

1 x 2 + y2

sec q

4 f = 0 Úq = 0 Úr = 0 p

p

= Ú 2 Ú 4 ÈÍ Ú 0 0 Î 0

sec q

x 2 + y2 + z2

r 2 sin q dr dq df r

r dr ˘˙ sin q dq df ˚

9.6

p 2

Ú Ú p

È

p 4

r2 2

p

Ú2 Ú4 1 2Ú

9.125

sec q

sin 0

˘ sec 2 q sin q q df 2

p

p 4

p

sec q q p

1 f 2 1 p 2 2 p 4

Triple Integrals

q 04 p - c0 4 -

)

Type II Evaluation of Triple Integrals Over the Given Region

Example 1 Evaluate Ú x 2 y over the region bounded by the planes x = 0, y = 0, z = 0 and x + y + z = 1. Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from xy-plane and terminates on the plane x + y + z = 1. z

y

R (0, 0, 1)

Q (0, 1) x+y

z

1 B

B Q (0, 1, 0)

O A

y

y

P (1, 0, 0) x

Fig. 9.131

O

A′

(1, 0) x

9.126

Chapter 9

Multiple Integrals

Limits of z : z = 0 to z = 1 – x – y 2. Projection of the plane x + y + z = 1 in xy-plane is DOPQ. Putting z = 0 in x + y + z = 1, the equation of the line PQ is obtained as x + y = 1. 3. Draw a vertical strip A¢B¢ in the region OPQ. A¢B¢ starts from the x-axis and terminates on the line x + y = 1. Limits of y : y = 0 to y = 1 – x Limits of x : x = 0 to x = 1 I=Ú

1 1- x 1- x - y 2 x yzz dz dy dx 0

Ú Ú

0 0

=Ú

1 1- x

Ú

0 0

x2 y

z2 2

1- x - y

dy dx 0

{

} }

1 1 2 È 1- x x y (1 - x )2 + y 2 - 2 y(1 - x ) dy ˘˙ dx ˚ 2 Ú0 ÍÎ Ú0 1 1 2 È 1- x = Ú x ÍÚ y(1 - x )2 + y3 - 2y 2 y 2 (1 - x ) dy ˘˙ dx ˚ 2 0 Î 0 =

{

=

1 1 2 y2 y 4 y3 x (1 - x )2 + - 2(1 - x ) Ú 2 0 2 4 3

1- x

dx 0

=

2 1 1 2È (1 - x ) 4 (1 - x )3 ˘ 2 (1 - x ) x ( 1 x ) ◊ + 2 ( 1 x ) ◊ Í ˙ dx 2 Ú0 Î 2 4 3 ˚

=

1 1 x2 (1 - x )4 dx 2 Ú0 12

(1 - x )7 1 (1 - x )5 2 (1 - x )6 = ◊x ◊ 2x + ◊2 -210 24 -5 30

1

0

1 Ê 1 ˆ = 0+ Á 24 Ë 105 ˜¯ =

1 2520

Example 2 Evaluate

ÚÚÚ 2xx dV ,

where E is the region under the plane

E

2x + 3y + z = 6 that lies in the first octant.

[Winter 2015]

Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from xy-plane and terminates on the plane 2x + 3y + z = 6. Limits of z : z = 0 to z = 6 – 2x – 3y

9.6

Triple Integrals

9.127

2. Projection of the plane 2x + 3y + z = 6 in xy-plane is DOPQ. Putting z = 0 in 2x + 3y + z = 6, the equation of the line PQ is obtained as 2x + 3y = 6. z

y

R (0, 0, 6)

Q (0, 2) B′ B Q (0, 2, 0)

O

y

A

O

A′

(3, 0) x

y

P (3, 0, 0) x

Fig. 9.132

3. Draw a vertical strip A¢B¢ in the region OPQ. A¢B¢ starts from the x-axis and terminates on the line 2x + 3y = 6. 6 - 2x 3

Limits of y : y = 0

to

y=

Limits of x : x = 0

to

x=3

I = ÚÚÚ 2 x dV

E 6 -2 x 3 6 -2 x -3 y 3 2 x dz dy dx 0 0 0

=Ú

Ú

Ú

6 -2 x 3 0 0

=Ú

3

Ú

6 -2 x 3 0 0

=2Ú

3

=2Ú

3

0

Ú

6 -2 x -3 y

2x z 0

dy dx

x(6 - 2 x - 3 y) dy dx

3 y2 x (6 - 2 x ) y 2

6 -2 x 3

dx 0

2 3 È (6 - 2 x ) 3 Ê 6 - 2 x ˆ ˘ ˙ dx = 2Ú x Í(6 - 2 x ) - Á 0 3 2 Ë 3 ˜¯ ˙ ÍÎ ˚

9.128

Chapter 9

Multiple Integrals

3

= 2Ú x 0

(6 - 2 x )2 dx 6

4 3 x (9 + x 2 - 6 x )dx 3 Ú0 4 3 = Ú (9 x + x 3 - 6 x 2 ) dx 3 0 =

x3 4 x2 x4 + -6 = 9 4 3 3 2

3

0

=9

Example 3 Evaluate ÚÚÚ xyz dx dy dz over the positive octant of the sphere x2 + y2 + z2 = 4. Solution Putting x = r sin q cos f, y = r sin q sin f, z = r cos q, the equation of the sphere x2 + y2 + z2 = 4 reduces to r2 sin2q cos2f + r2 sin2q sin2f + r2 cos2 q = 4, r2 = 4, r = 2.

z

The region is the positive octant of the sphere r = 2. Limits of r : r = 0 to r = 2 Limits of

q=0

to

θ=

Limits of

f=0

to

φ=

π 2 π

r

P

q y

O φ

x

2

Fig. 9.133

Hence, the spherical form of the given integral is I = ÚÚÚ xyz dx dy dz p

p

0

0

2

= Ú 2 Ú 2 Ú (r 3 sin 2 q cos q cos f si sin n f )r 2 sin q dr dq df 0

p 2 0

p

= Ú sinn 3 q cos cos q d dq qÚ 2 sinn 4 q = 4

0

p 2 0

- cos 2f 4

p 2 0

2 sin 2f df Ú r 5 dr 0 2

r6 6

2

0

È ˘ [ f (q )]n +1 n , n π 1˙ Í∵ Ú [ f (q )] f ¢(q )dq = n +1 Î ˚

9.6

=

1Ê 4 p ˆÈ 1 ˘ Ê 26 ˆ sin - sin 0˜ Í - (cos p - coss 0)˙ Á ˜ Á ¯Î 4 4Ë 2 ˚Ë 6 ¯

=

4 3

Triple Integrals

9.129

Example 4 Evaluate

dx dy dz

ÚÚÚÚ

over the region bounded by the sphere

a2 - x 2 - y2 - z2 x2 + y2 + z2 = a2. Solution

1. Putting x = r sin q cos f, y = r sin q sin f, z = r cos q, the equation of the sphere x2 + y2 + z2 = a2 reduces to r = a. 2. For the complete sphere, limits of r : r = 0 to r = a limits of q : q = 0 to q = p limits of f : f = 0 to f = 2p Hence, the spherical form of the given integral is dx dy dz

I = ÚÚÚ =Ú

2p

=Ú

2p

a - x 2 - y2 - z2 2

p

Ú0 Ú0

0

0

a r2

sin q dr dq df a2 - r 2

p

a

0

0

df d f Ú sin q dq Ú

r 2 + a2 - a2 a2 - r 2

dr

ˆ aÊ a2 2p p = f 0 ◊ - cos q 0 ◊ Ú Á - a 2 - r 2 ˜ dr 0 Ë a2 - r 2 ¯ = (22p p )(- cos p + cos 0) a sin 2

-1

Ê ˆ a2 = 2p (2) Á a 2 sin sin -1 1 sin -1 1˜ Ë ¯ 2 Ê a2 ˆ = 4p Á sin -1 1˜ Ë 2 ¯ = 4p ◊

a2 p ◊ 2 2

= p 2 a2

r r a2 -1 r a2 - r 2 sin -1 a 2 2 a

a

0

9.130

Chapter 9

Multiple Integrals

Example 5 Evaluate

d

∫∫∫

over the region bounded by the spheres

1 2

( z ) 2 2 x + y + z = a and x + y 2 + z 2 = b2, a > b > 0. 2

2

2

z

Solution 1. Putting x = r sinq cos f, y = r sinq sin f, z = r cosq, equations of the spheres x2 + y2 + z2 = a2 and x 2 + y 2 + z 2 = b2 reduce to r = a and r = b respectively. 2. Draw an elementary radius vector OAB from the origin in the region. This radius vector enters in the region from the sphere r = b and terminates on the sphere r = a. 3. Limits of r : r = b to r = a. For the complete sphere, limits of q : q = 0 to q = p limits of f : f = 0 to f = 2p Hence, the spherical form of the given integral is dx y dz I ÚÚÚ p

Ú Ú Ú 2p

Ú0

a

a b y

x

Fig. 9.134 1

r 2 sinq dr dq df r a

dq Ú r r

d

b

p 0

◊2p (

O

r r

+ z )2

(x 2p

A

+ 2

2

r ◊ 2 0)

a

b

(a 2

)

).

Example 6 Evaluate

ÚÚÚ z

2

x dy z over the region common to the sphere

x2 + y2 + z2 = 4 and the cylinder x2 + y2 = 2x.

9.6

Triple Integrals

9.131

Solution 1. Putting x = r cos q, y = r sin q, z = z, the equation of (i) the sphere x 2 + y 2 + z 2 = 4 reduces to r2 + z2 = 4 z 2 = 4 – r 2.

q

A

(ii) the cylinder x2 + y 2 = 2x reduces to r 2 = 2r cosq, r = 2 cosq. 2. Draw an elementary volume parallel to z-axis in the region. This elementary volume starts from the part of the sphere z2 = 4 – r 2, below xyplane and terminates on the part of the sphere z2 = 4 – r 2, above xy-plane. Limits of r z = -

p 2

r = 2cos q O

q=0

Fig. 9.135

r

3. Projection of the region in rq -plane is the circle r = 2 cosq. 4. Draw an elementary radius vector OA in the region (r = 2 cosq ) which starts from the origin and terminates on the circle r = 2 cosq Limits of r : r = 0 to r = 2 cos q p p Limits of q q to q = 2 2 Hence, the cylindrical form of the given integral is z 2 dx y dz

I p

= Ú -2p Ú

2 cosq

Ú

4 r

z2r d

4 r2

r dq

2

=Ú

p 2 p 2

1 = 3 =

1 3

Ú

2 cosq

p

r r dq -r 3

2 cosq

p

2 cosq

p 2

1 3 Ú-

r dr dq

r

p 2

3

-(4 r ) 2 ( 2r r dq 5

p

=-

4 -r

z3 3

2

q

-

Ú

2

0

p

2 = - Ú 2p È 15 - (4 2

5

c

2

5

q ) - ( 4) 2

dq

(r

)dr

[ (r )]n +1 ˘ +1

9.132

Chapter 9

Multiple Integrals

p

=-

2 2 5 5 (2 sin q - 25 ) dq 15 Ú- p 2

2 È = - Í0 - 2 5 q 15 Í Î

È∵ a f (q )dq = 0, iiff f (-q ) = - f (q )˘ Í Ú- a ˙ Í ˙ 5 5 -q q ) = - sin sin q ÎHere sinn (˚

˘ ˙ p - ˙ 2˚ p 2

26 p 15 64p = 15 =

Example 7 Evaluate ÚÚÚÚ xyz dx dy dz over the region bounded by the planes x = 0, y = 0, z = 0, z = 1 and the cylinder x2 + y 2 = 1. Solution 1. Putting x = r cosq, y = r sinq, z = z, equation of the cylinder x2 + y2 = 1 reduces to r 2 = 1, r = 1. z

B

q= p 2

z=1

A A′ r=1

O

y

O

q=0

A

x

Fig. 9.136

2. Draw an elementary volume AB parallel to z-axis in the region. This elementary volume AB starts from xy-plane and terminates on the plane z = 1. Limits of z : z = 0 to z = 1

9.6

Triple Integrals

9.133

3. Projection of the region in rq -plane is the part of the circle r = 1 in the first quadrant. 4. Draw an elementary radius vector OA¢ in the region in the rq -plane which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0

to

r=1 p Limits of q : q = 0 to q = 2 Hence, the cylindrical form of the given integral is I = ÚÚÚ xyz dx dy dz =Ú

p 1 2 r2 z=0 q =0 r =0 1

Ú Ú p

1

= Ú z dz Ú 2 0

z2 = 2

0

1

0

cos q sin q ◊ zzrr dz dr dq

1 sin si n 2q dq Ú r 3 dr 0 2

cos 2q 4

p 2 0

r4 4

1

0

1 = 16

Example 8 Evaluate

∫∫∫∫

x 2 + y 2 dx dy dz dz over the region bounded by the right

circular cone x2 + y2 = z2, z > 0 and the planes z = 0 and z = 1. Solution 1. Putting x = r cosq, y = r sinq, z = z, the equation of the cone x 2 + y 2 = z 2 reduces to r 2 = z 2, r = z. 2. Draw an elementary volume AB parallel to z-axis in the region, which starts from the cone r = z and terminates on the plane z = 1. Limits of z : z = r to z = 1. 3. Projection of the region in rq -plane is the curve of intersection of the cone r = z and the plane z = 1 which is obtained as r = 1, a circle with centre at the origin and radius 1. 4. Draw an elementary radius vector OA¢ in the region which starts from the origin and terminates on the circle r = 1. Limits of r : r = 0 to r = 1 Limits of q : q = 0 to q = 2p

9.134

Chapter 9

Multiple Integrals

z

z

B r

1 = p 2

z A′

A

r y

O

1 =0

x

Fig. 9.137

Hence, the cylindrical form of the given integral is I

dx 2p

1

1

◊r d

=0 r =0 z =r 2p

=

Ú

1 2

1

1

r3 r4 − 3 4

0

⋅

r dq

r z r d dq

2p

2

dz

- dr 1

1 12

6

Example 9 Evaluate Ú ( ) dz over the region bounded by the parabo2 2 loid x + y = 3z and the plane z = 3. Solution 1. Putting x = r cos q, y = r sin q, z = z, the equation of the paraboloid x2 + y2 = 3z reduces to r 2 = 3z. 2. Draw an elementary volume AB parallel to z-axis in the region which starts from the paraboloid r 2 = 3z and terminates on the plane z = 3.

9.6

Triple Integrals

9.135

r2 to z = 3 3 3. Projection of the region in rq -plane is the curve of intersection of the paraboloid r 2 = 3z and the plane z = 3 which is obtained as r 2 = 9, r = 3, a circle with centre at the origin and radius 1. Limits of z : z =

z

q= p 2

z=3

A′

r 2 = 3z A

q y

O

O

r=3 q=0

x

Fig. 9.138

4. Draw an elementary radius vector OA¢ in the region (circle r = 3) which starts from origin and terminates on the circle r = 3. Limits of r : r = 0 to r = 3 Limits of q : q = 0 to q = 2p Hence, the cylindrical form of the given integral is I = ÚÚÚ ( x 2 + y 2 )dx dy dz =Ú

2p

=Ú

2p

=Ú

2p

0

0

0

3 3

Ú0 Ú r

3 3 3 r 0

Ú

r 2 ◊ r dz dr dq z

3 r2 3

dr dq

Ê 3 r2 ˆ dq d q ◊ Ú r 3 Á 3 - ˜ dr 0 3¯ Ë

2p

=q0

2

3r 4 r 6 4 18

Ê3 3 ˆ = 2p Á - ˜ Ë 4 18 ¯ 81p = 2 5

6

3

0

9.136

Chapter 9

Multiple Integrals

Example 10 Evaluate

ÚÚÚ V

ellipsoid

1-

x 2 y2 z2 - dx dy dz, where V is the volume of the a 2 b2 c 2

x 2 y2 z2 + + = 1. a 2 b2 c 2

Solution It is difficult to integrate this integral in cartesian form. Therefore, transforming the ellipsoid into a sphere using following change of variables. x 2 y2 z2 x y z = u, = v, = w, equation of the ellipsoid 2 + 2 + 2 = 1 reduces to a b c a b c u 2 + v 2 + w2 = 1, which is a sphere of radius 1 and centre at the origin,

Putting

dx dy dz = |J| du dv dw

where,

∂x ∂u ∂( x, y, y, z ) ∂y J= = ∂(u, v, v, w) ∂u ∂z ∂u

∂x ∂v ∂y ∂v ∂z ∂v

∂x ∂w ∂y ∂w ∂z ∂w

a 0 0 = 0 b 0 = abc 0 0 c Therefore, dx dy d dzz = abc du dv dw New form of the integral is I = Ú Ú Ú 1-

x2 a

2

-

y2 b

2

-

z2 c2

dx dy dz

= Ú Ú Ú 1 - u2 - v 2 - w 2 ◊ abc du dv dw Since in the new coordinate system u, v, w, the region of integration is a sphere, therefore using spherical coordinates u = r sinq cos f, v = r sin q sin f, w = r cosq and du dv dw = r2 sin q dr dq df, the equation of the sphere u2 + v2 + w2 = 1 reduce to r2 =1, r = 1. For complete sphere limits of r : r = 0 to r = 1 (radius of sphere) limits of θ : θ = 0 to θ = π limits of φ : φ = 0 to φ = 2π

9.6

Triple Integrals

9.137

Hence, the spherical form of the given integral is I=Ú

2p

0

p

1

2p

df Ú sin q dq Ú r 2 1 - r 2 dr

Ú0 Ú0

= abc Ú

0

1 - r 2 abc ◊ r 2 sin q dr dq df p

1

0

0

ddrr = cos t dt

Putting r = sin t, When r = 0,

t=0

When r = 1,

t=

p 2 2p

p

\ I = abc f 0 - cos q 0

p

Ú02 sin

1 1 π = abc(2π )(2) ⋅  ⋅ ⋅   4 2 2 =

2

t ◊ cos t ◊ cos t dt

[Using reduction formula]

π 2 abc 4

Example 11 Evaluate ÚÚÚÚ x 2 y 2 z 2 dx dy dz over the region bounded by the surfaces xy = 4, xy = 9, yz = 1, yz = 4, zx = 25, zx = 49. Solution Evaluation of integral becomes easier by changing the variables. Under the transformation xy = u, yz = v, zx = w, the surfaces get transformed to u = 4, u = 9, v = 1, v = 4, w = 25, w = 49. These equations represent the planes parallel to v w, w u and uv planes in the new coordinate system. It is easier to find partial derivatives of u, v, w w.r.t. x, y and z.

∂u ∂x

∂u ∂y

∂u ∂z

∂(u, v, w) ∂v = ∂( x, y, z ) ∂x ∂w ∂x

∂v ∂y ∂w ∂y

∂v ∂z ∂w ∂z

y = 0 z

x 0 z y 0

x

= y( zzxx - 0) - x(0 - yz ) = 2 xyz

9.138

Chapter 9

Multiple Integrals

du dv ddw w = J dx ddyy dz = 2 xyz ddxx dy dz

dx dy d dzz =

1 du dv dw 2 xyz 1

=

[∵ x 2 y 2 z 2 = uvw]

du dv dw 2 uvw Limits of u : u = 4 to u = 9 Limits of v : v = 1 to v = 4 Limits of w : w = 25 to w = 49 Hence, the new form of the integral is

I = ÚÚÚ x 2 y 2 z 2 dx dy dz =Ú

49

4

9

uvw ◊ w = 25 Úv =1 Úu = 4 2 1

=

1

1 uvw

du dv dw

1

4 9 1 49 2 w dw Ú v 2 dv Ú u 2 du Ú 1 4 2 25 3 2w 2

49

3 2v 2

4

3 2u 2

9

1 2 3 25 3 1 3 4 4 = (343 -125 - 125)(8 - 1)(27 - 8) 27 115976 = 27 =

EXERCISE 9.7 (I) Evaluate the following integrals: 1.

2.

3.

1

2

0

0

∫ dx ∫

2

dy ∫ x 2 yyzz dz 1

log 2

x

x +y

0

0

0

∫ ∫ ∫ π 2 0

∫ ∫

a cos θ

0

Ans. :1

e x + y + z dz dy d x 5   Ans. : 8   

∫

0

a2 −r 2

r dz dr d θ

 a3  π 2    Ans. :  −   3 2 3  

9.6

4.

π

a(1+ 1+ cos θ )

0

0

∫ ∫

∫

h

0

Triple Integrals

9.139

  r 2 1 − r dz dr d θ 1 + cos θ)   a(1+  π a 2h   Ans. :  2  

5.

4

2 z

0

0

4 z−x2

∫∫ ∫

0

dy dx dz Ans. : 8π 

6.

7.

π 2 0

∫ ∫

a sin θ

0

2

a −r a

∫

0

2

y

x +y

0

0

x −y

∫∫ ∫

2

r dz dr dθ

 5a 3   Ans. :  64  

(x + y + z)dz dx dy Ans. :16 

8.

a

∫∫ 0

0

a2 − x 2

∫

a2 − x 2 − y 2

0

xyzz dz dy dx.  a6   Ans. :  48  

(II) Evaluate the following integrals over the given region of integration: 1.

∫∫∫ (x + y + z)dx dy dz

over the tetrahedron bounded by the planes

x = 0, y = 0, z = 0 and x + y + z = 1. 1   Ans. : 8    2.

dx dy dz

∫∫∫ (1 + x + y + z)

3

over the tetrahedron bounded by the planes x = 0,

y = 0, z = 0 and x + y + z = 1.

3.

∫∫∫ xyzz dx dy dz

 1 5   Ans. :  log 2 −   2 8  

over the positive octant of the sphere x2 + y2 + z2 = a2.  a6   Ans. :  48  

4.

∫∫∫ xyzz(x

2

+ y 2 + z 2 )dx dy dz over the positive octant of the sphere

x 2 + y 2 + z 2 = a2.

 a8  Ans. :   64  

9.140

5.

Chapter 9

∫∫∫ ((yy

2

Multiple Integrals

z 2 + z 2 x 2 + x 2 y 2 ) dx dy dz over the sphere of radius a and centre

at the origin.

6.

7.

z2 ∫∫∫ x 2 + y 2 + z 2 dx dy dz over the sphere x 2 + y 2 + z 2 = 2.

dx dy dz

∫∫∫

3 2 2

 4π a 7  Ans. :   35    8π 2   Ans. :  9  

over the region bounded by the spheres x2 + y2 + z2 = a2

(x 2 + y 2 + z ) and x2 + y 2 + z2 = b2, a > b > 0.   a  Ans. : 4π log    b   8.

∫∫∫ z dx dy dz 2

over the region common to the spheres x 2 + y 2 + z 2 = a 2

and cylinder x 2 + y 2 = ax.

9.

∫∫∫ (x

2

+ y 2 )dx dy dz over the region bounded by the paraboloid x2 + y2 = 2z

and the plane z = 2.

∫∫∫ x

 2π a 5   Ans. :  15  

16π    Ans. : 3   

2

10.

y z dx dy dz over the tetrahedron bounded by the planes x = 0, x y z y = 0, z = 0 and + + = 1. a b c  a 3 b 2c 2  Ans. :   2520  

11.

∫∫∫ xyzz dx dy dz 2

2

over the positive octant of the ellipsoid

2

x y z + 2 + 2 ≤ 1. 2 a b c

12.

 a 2 b 2c 2  Ans. :   48  

x 2 y 2 z2 ∫∫∫ 1 + 4 + 9 dx dy dz over the region bounded by the ellipsoid x 2 y 2 z2 + + = 1. 1 4 9 Ans. : 8π 

9.7 Area as Double Integral

9.141

9.7 AREA AS DOUBLE INTEGRAL 9.7.1 Area in Cartesian Coordinates (i) The area A bounded by the curves y = y1(x) and y = y2(x) intersecting at the points P (a, b) and Q (c, d ) is

y

Q c, d )

x (ii) If equation of the curves are represented as x = x1(y) and x = x2(y) then d

P (a b)

y

Note: Consider the symmetricity of the region while calculating area.

x

O

Fig. 9.139

Example 1 Find the area bounded by the ellipse

x2 y2 + = 1, above x-axis. a2 b

Solution 1. The region is symmetric about y-axis. Total area = 2 (area bounded by the ellipse in the first quadrant) 2. Draw a vertical strip AB in the region which lies in the first quadrant. AB starts from the x-axis and terminates on the 2 x2 ellipse 2 + =1 a b x2 Limits of y : y = 0 to a2 Limits of x : x = 0 to x = a a

x

b 1

a

2b a

2

− a

O

B

A

Fig. 9.140

x

= 2∫ y 0 =

(0, b)

x

b 1

a

y

2

dx

x2 dx a2 2

x

2 x +y 1 b2 a

x P (a, 0)

9.142

Chapter 9

Multiple Integrals

a

2b x 2 a2 x a − x 2 + sin −1 a 2 2 a0 2b  a 2 −1  =  sin 1 a  2 =

2b  a 2 π   ⋅  a  2 2 π ab = 2 =

Example 2 Find the area bounded by the parabola y 2 = 4x and the line 2x –3y + 4 = 0. Solution 1. The points of intersection of the parabola y2 = 4x and the line 2x – 3y + 4 = 0 are obtained as 2  2x + 4   = 4 x 3 y ( x + 2) 2 = 9 x x2 − 5x + 4 = 0 x = 1, 4

B

∴ y = 2, 4

A

The points of intersection are P (1, 2) and Q (4, 4). 2. Draw a vertical strip AB which starts from the line 2x – 3y + 4 = 0 and terminates on the parabola y2 = 4x. 2x + 4 to y = 2 x Limits of y : y = 3 Limits of x : x = 1 to x = 4 A = 2∫

4

1

4

∫

2 x 2 x+4 3

2xx – 3yy + 4 = 0

=∫

4

1

x

O y 2 = 4x

Fig. 9.141

2 x 3

2x + 4  dx  2 x −  dx 3  3 2

P (1, 2)

dy dx dx

= ∫ y 2 x+ dx x + 4 dx 1

Q (4, 4)

4 2

= 2 ⋅ 2 ⋅ x − x − 4x 3 3 3 1 4 1 4 = (8 − 1) − (16 −1 − 1) − (4 − 1) 3 3 3 1 = 3

9.7 Area as Double Integral

9.143

Example 3 Find the area enclosed by the curves y = x2 and y = x. Solution y

1. The points of intersection of the parabola y = x2 and the line y = x are obtained as x = x2 x = 0, 1 \ y = 0, 1 The points of intersection are O (0, 0) and P (1, 1). 2. Draw a vertical strip AB which starts from the parabola y = x2 and terminates on the line y = x. Limits of y : y = x2 to y = x Limits of x : x = 0 to x = 1 A=∫

1

0

∫

x x2

y =x

P(1,1) y = x2

B A O

x

Fig. 9.142

dy dx dx

1

x

0

x

= ∫ y 2 dx 1

= ∫ ( x − x 2 ) dx dx 0

x 2 x3 = − 2 3

1

0

1 1 = − 2 3 1 = 6

Example 4 Find the area enclosed by the parabola y2 = 4ax and the lines x + y = 3a, y = 0 in the first quadrant. Solution 1. The points of intersection of the parabola y2 = 4ax and the line x + y = 3a are obtained as y2 = 4a(3a – y) 2 y + 4ay – 12a2 = 0 y = 2a, – 6a \ x = a, 9a The point of intersection is Q (a, 2a) which lies in the first quadrant. 2. Area enclosed in the first quadrant is OPQ.

9.144

Chapter 9

Multiple Integrals

Draw a horizontal strip AB which starts from the parabola y2 = 4ax and terminates on the line x + y = 3a. y y2 4a Limits of y : y = 0

Limits of x x

2a

=∫

to x = 3a – y to

3

x

y 2 = 4ax

Q(a 2a)

y

4a 3a y 2a

B

y

P x

O

4a 2

− y2 − 2

=

4

y2 4a

y y3 3

1 8a − ⋅ 4a 3 =

x + y = 3a

y = 2a

2a

0

Fig. 9.143

3

10 2 a 3

Note: In case of vertical strip, two vertical strips are required to cover the entire region. Therefore one horizontal strip is preferred over vertical strip.

Example 5 Find the area bounded by the parabolas y2 = 4ax and x2 = 4ay. Solution 1. The points of intersection of the parabolas y2 = 4ax and x2 = 4ay are obtained as x2 4a

2

= 4ax

x4 = 16a2 (4ax) x(x3 – 64a3) = 0 x = 0, x = 4a \ y = 0, y = 4a The points of intersection are O (0, 0) and P (4a, 4a). 2. Draw a vertical strip AB which starts from the parabola x2 = 4ay and terminates on the parabola y2 = 4ax. x2 ax Limits of y y = to y 4a Limits of x : x = 0 to x = 4a

y y = 4ax B

P(4a a)

x 2 = 4ay A O

Fig. 9.144

x

9.7 Area as Double Integral

4a

A=∫

0

=∫

4a

0

∫

2 ax x2 4a

y

dy dx dx

2 ax x2 4a

9.145

dx

4a  x2  = ∫  2 ax −  dx dx 0  4a  4a

2 3 1 x3 = 2 a ⋅ x2 − ⋅ 3 4a 3

0

3 4 1 = a ( 4a ) 2 − (4a 4a )3 3 12a 32 2 16 2 = a − a 3 3 16 = a2 3

Example 6 Find the area enclosed by the curves y = 2 – x and y2 = 2(2 – x). Solution 1. The points of intersection of the line y = 2 – x and the parabola y2 = 2(2 – x) are obtained as y (2 – x)2 = 2(2 – x) (2 – x)(2 – x – 2) = 0 (2 – x)(–x) = 0 x = 2, 0 y = 0, 2 The points of intersection are P (2, 0) and Q (0, 2). 2. Draw a vertical strip AB which starts from the line y = 2 – x and terminates on the parabola y2 = 2 (2 – x).

Q (0, 2) y =2−x

Limits of y : y = 2 – x to y = 2(2 − x) Limits of x : x = 0 to x = 2 A=∫

2

0 2

∫

2− x

=∫ y 0

2( 2− x )

dy dx dx

2( 2− x ) 2− x

dx

1 2  = ∫  2 (2 − x) 2 − (2 − x)  dx dx 0  

B y 2 = 2 (2 −x) A O

Fig. 9.145

P (2, 0) x

9.146

Chapter 9

Multiple Integrals

2

3

2(2 − x) 2 x2 2⋅ − 2x + −3 2

=

0

8 1  =  0 +  − 2( 2 − 0 ) + ( 4 − 0 )   3 2 =

2 3

Example 7 Find the area bounded between the parabolas x2 = 4ay and x2 = – 4a (y – 2a). Solution 1. The parabola x2 = 4ay has vertex (0, 0) and the parabola x2 = – 4a(y – 2a) has vertex (0, 2a). Both the parabolas are y symmetric about the y-axis. 2. The points of intersection of Q (0, 2a) x2 = 4ay and x2 = – 4a(y – 2a) are B x 2 = 4ay obtained as 4ay = −4 4a a ( y − 2a ) R P (2a, a) (−2a, a)

8ay = 8a 2 y=a

x 2 = −4a (y − 2a) O

A

x

∴ x = ± 2a The points of intersection are P (2a, a) and R (–2a, a). 3. The region is symmetric about y-axis. Total area = 2 (Area in the first quadrant)

Fig. 9.146

4. Draw a vertical strip AB in the region which lies in the first quadrant. AB starts from the parabola x2 = 4ay and terminates on the parabola x2 = – 4a(y – 2a). x2 x2 Limit of y : y = to y = 2a − 4a 4a Limits of x : x = 0 to x = 2a A = 2∫

2a

= 2∫

2a

0

0

∫

2a −

x2 4a

x2 4a

y

2a − x2 4a

dy dx dx

x2 4a

dx dx

2a  x2 x2  = 2∫  2 a − d −  dx 0  4a 4a 

9.7 Area as Double Integral

9.147

2a

x3 6a 0 4   = 2  4a 2 − a 2   3  16 = a2 3 = 2 2ax −

Example 8 Find smaller of the area enclosed by the curves y = 2 – x and x2 + y2 = 4. Solution 1. The points of intersection of the line y = 2 – x and the circle x2 + y2 = 4 are obtained as x2 + (2 – x)2 = 4 x + 4 – 4x + x2 = 4 2x2 = 4x x = 2, 0 \ y = 0, 2

y

2

Q (0, 2) B y

x2 + y2 = 4

=

The points of intersection are P (2, 0) and Q (0, 2). 2. Draw a vertical strip AB which starts from the line y = 2 – x and terminates on the circle x2 + y2 = 4.

2 − x

A O

Limits of y : y = 2 – x to y = 4 − x 2 Limits of x : x = 0

to x = 2

A=∫

2

0 2

∫

4 − x2 2− x

=∫ y 0

4− x 2− x

dy dx dx

Fig. 9.147

2

dx

= ∫  4 − x 2 − (2 − x)  dx dx 0   2

x 4 x x2 = 4 − x 2 + sin −1 − 2 x + 2 2 2 2

2

0

= 2 sin −1 1 − 2 =π −2

Example 9 Find the area of the loop of the curve x(x2 + y2) = a (x2 – y2).

P (2, 0) x

9.148

Chapter 9

Multiple Integrals

Solution

 a − x The equation of the curve can be rewritten as y 2 = x 2   a + x  1. The points of intersection of the curve with x-axis ( y = 0) are obtained as x2(a – x) = 0 x = 0, x = a. y The loop of the curve lies between the points O (0, 0) and P (a, 0). 2. The region is symmetric about x-axis B Total area = 2 (Area above x-axis) P (a, 0) 3. Draw a vertical strip AB in the region x A O above x-axis. AB starts from x-axis and x=−a a − x terminates on the curve y 2 = x 2  .  a + x Limits of y : y = 0 to

y=x

Limits of x : x = 0

x=a

to

A = 2∫

a

0

∫

a− x a+ x

x

0

a

x

= 2∫ y 0 0

a

= 2∫ x 0

Putting When When

a−x a+ x

dyy dx d

a−x a+ x

dx dx

a−x dx dx a+ x

x = a cosq, dx = – a sin q dq π x = 0, θ = 2 x = a, q = 0 0

A = 2 Ú p a cos q 2

a - a cos q (- a sin q )dq a + a cos q

q 2 dq = 2 a Ú cos q sin q ◊ q 2 co oss2 2 q p sin q q 2 2 2 dq = 2 a Ú cos cos q ◊ 2 ssin cos ◊ 0 q 2 2 cos 2 2

p 2 0

p

2 sin 2

= 2 a 2 Ú 2 coss q (1 - coss q ) dq 0

Fig. 9.148

9.149

9.7 Area as Double Integral

p

1 + cos 2q ˆ Ê = 2 a 2 Ú 2 Á cos q ˜¯ dq 0 Ë 2 q sin sin 2q = 2 a sin q - 2 4

p 2

2

0

ÈÊ p ˘ ˆ 1Êp ˆ 1 = 2 a 2 ÍÁ sin - sin n 0˜ - Á - 0˜ - (sin p - sin 0 )˙ Ë ¯ Ë ¯ 2 2 2 4 Î ˚ p Ê ˆ = 2a 2 Á 1 - ˜ Ë 4¯

Example 10 Find the area included between the curve y2(2a – x) = x3 and its asymptote. [Summer 2017] Solution The equation of the curve can be rewritten as x3 2a - x 1. The point of intersection of the curve with x-axis (y = 0) is x = 0. 2. The region is symmetric about x-axis. Total area = 2 (Area above x-axis) 3. Draw a vertical strip AB in the region above x-axis. AB starts from x-axis and terminates on the curve x3 . y2 = 2a - x

y

y2 =

Limits of y : y = 0 to y = x

x 2a - x

Limits of x : x = 0 to x = 2 a A = 2Ú

2a

= 2Ú

2a

= 2Ú

2a

0

x

Ú0

x

y

0

0

x 2a - x

x

0

dy dx

x 2a - x

dx

x dx 2a - x

B

O

A

Fig. 9.149

Asymptote

x x = 2a

9.150

Chapter 9

Multiple Integrals

x = 2a sin2q, dx = 2a (2 sinq cosq dq) When x = 0, q = 0 p When x = 2 a, q = 2

Putting

p

2 a sin 2 q

0

2 a cos2 q

A = 2 Ú 2 2 a sin 2 q p

= 2 Ú 2 2 a sin 2 q ◊ 0

◊ 4 a sin q cos q dq

sin q ◊ 4 a sin q cos q dq cos q

p

= 16 a 2 Ú 2 sin 4 q dq 0

= 16 a 2 ◊

3 1 p ◊ ◊ 4 4 4

= 3p a 2

Example 11 Find the area between the rectangular hyperbola 3xy = 2 and the line 12x + y = 6. Solution 1. The points of intersection of the rectangular hyperbola 3xy = 2 and the line 12x + y = 6 are obtained as y 3 x (6 − 12 x) = 2

18 x 2 − 9 x + 1 = 0

Q

1 , 4 6

1 1 x= , 3 6 ∴ y = 2, 4 The points of intersection are 1  1  P  , 2 and Q  , 4 . 3  6  2. Draw a vertical strip AB in the region which starts from the rectangular hyperbola 3xy = 2 and terminates on the line 12x + y = 6. 2 Limits of y : y = to y = 6 – 12x 3x 1 1 Limits of x : x = to x = 6 3

12x + y = 6 B 3xy = 2

A

P 1, 2 3 O

x

Fig. 9.150

9.151

9.7 Area as Double Integral

1

6

2x

x

3x

1

= ∫ 13 y 6

6 12 x 2 3x

dx

1

2 dx 3x

= ∫ 13 6 12 x 6

1

=

3 2 log x 1 3

2

6

2 1 1 1 − log − log 3 36 3 6

1 −6 9 =

1 2 − log 2 2 3

Example 12 Find the area bounded by the hypocycloid x a Solution 1. The hypocycloid is symmetric in all the quadrants. Total area = 4 (area in the first quadrant) 2. Draw a vertical strip AB parallel to y-axis in the region which lies in the first quadrant. AB starts from x-axis and terminates on the curve

x a

2 3

2 3

y b

+

x a

Limits of x : x = 0

to x = a

4∫

a

 b1

2 2

x

x x a

3 2 2

3

x a

2 3

1.

y Q (0, b) x a B

2 3

A

= 1.

to

b 1−

y + b

O

Limits of y : y = 0

a

2 3

2 3

2 dx

2 3

3 2

Fig. 9.151

y b

2 3

P (a

=

x

9.152

Putting When When

Chapter 9

Multiple Integrals

x = a cos3 t, dx = 3a cos2 t (–sin t) dt

π 2 x = a, t = 0 x = 0, t =

3

0

A = 4∫π b(1 − cos 2 t ) 2 ( −3a coss 2 t sin t) t dt 2

π

= 12ab∫ 2 sin n 4 t cos 2 t dt 0

3 1 1 π = 12ab ⋅ ⋅ ⋅ 6 4 2 2 3 = π ab 8

[ Using reduction for ormu mula ]

EXERCISE 9.8 1. Find the area bounded by y-axis, the line y = 2x and the line y = 4. [Ans. : 4] 2. Find the area bounded by the lines y = 2 + x, y = 2 – x and x = 5. [Ans. : 25] 3. Find the area bounded by the parabola y2 + x = 0, and the line y = x + 2. 9   Ans. : 2    4. Find the area bounded by the parabola x = y – y 2 and the line x + y = 0. 4   Ans. : 3    5. Find the area bounded by the curves y 2 = 4x and 2x – 3y + 4 = 0. 1   Ans. : 3    6. Find the area bounded by the parabola y = x2 – 3x and the line y = 2x. 125    Ans. : 6   

9.7 Area as Double Integral

9.153

7. Find the area bounded by the parabolas y2 = x, x2 = –8y. 8   Ans. : 3    8. Find the area bounded by the parabolas y = ax2 and y = 1 − a > 0.

x2 , where a

 4 a   Ans. :  2 3 a + 1    a + x 9. Find the area of the loop of the curve y 2 = x 2   a − x    2 π  Ans. : 2a  − 1  4   10. Find the area of one of the loops of x4 + y4 = 2a2xy.  π a2  Ans. :   4   11. Find the area enclosed by the curve 9xy = 4 and the line 2x + y = 2. 1 4    Ans. : 3 − 9 log 2   12. Find the area of the smaller region bounded by the circle x2 + y2 = 9 and a straight line x = 3 – y.   2π 3 −  Ans. : 4   2    3  13. Find the area bounded by the x-axis, circle x2 + y2 = 16 and the line y = x. [Ans. : 2p ] 14. Find the area bounded between the curves y = 3x2 – x – 3 and y = – 2x2 + 4x + 7. 45    Ans. : 2    2

2

2

15. Find the area bounded by the asteroid (x) 3 + (y ) 3 = (a) 3 . 3 2   Ans. : 8 π a   

9.154

Chapter 9

Multiple Integrals

9.7.2 Area in Polar Coordinates

p 2

q =

The area A bounded by the curves r = r1 (q ), r = r2 (q ) and the lines q = q1 and q = q2 is A=∫

θ2

θ1

∫

r2 (θ )

r1 (θ )

q = q2 R q = q1

S

r dr dθ

q2

Note: Consider the symmetricity of the region while calculating the area.

q1 P

Q r = r2(q ) r = r1(q )

O

q =0

Fig. 9.152

Example 1 Find the area between the circles r = 2 sinq and r = 4 sinq. Solution 1. The region is symmetric about the line θ = Total area = 2 (area in the first quadrant) 2. Draw an elementary radius vector OAB from the origin in the region which lies in the first quadrant. OAB enters in the region from the circle r = 2sinq and terminates on at the circle r = 4 sinq. Limits of r : r = 2sinq to r = 4sinq π Limits of θ : θ = 0 to θ = 2 π

A = 2∫ 2 ∫ 0

π

= 2∫ 2 0

4 sin θ 2 sin θ

r2 2

π . 2

P B r = 4 sinq Q

A

r dr dθ

r = 2 sinq q

4 sin θ

O

dθ 2 sin θ

π 2 0

= ∫ (16 sin 2 θ − 4 sin sin 2 θ ) dθ π

= ∫ 2 12 sin 2 θ dθ 0

π

= ∫ 2 6(1 − cos 2θ ) dθ 0

ssiin 2θ =6θ− 2

π 2

dθ

0

sin π − ssin 0   π sin = 6 −  2 2 = 3π

q= p 2

Fig. 9.153

q =0

9.7 Area as Double Integral

9.155

Example 2 Use double integral in polar form to find the area enclosed by the three petalled rose r = sin 3q. [Winter 2015] Solution 1. This curve consists of three similar loops. Total area = 3 (area of the loop in the first quadrant) 2. When r = 0, sin 3q = 0 3q = 0, p, 2p, 3p, ... p 2p , p , ... q = 0, , 3 3 Since, in the first quadrant, p r = 0 at q = 0, Fig. 9.154 3 p loop exists between q = 0 and q = . 3 3. Draw an elementary radius vector OA from the origin in the loop which lies in the first quadrant. OA starts from the origin and terminates on the curve = sin q. Limits of r : r = 0 to r = sin 3q Limits of q : q = 0 to q =

p 3 p

A = 3Ú 3 Ú

sin 3q

0

0

p

r2 2

= 3Ú 3 0

r dr dq

sin 3q

dq 0

p 3 0

=

3 sin 2 3q dq 2Ú

=

3 3 Ê 1 - coss 6q ˆ ˜¯ dq 2 Ú0 ÁË 2

p

sin 6q 3 = q4 6

p 3 0

3Êp 1 ˆ = Á - sin 2p ˜ Ë ¯ 4 3 6 =

3Êpˆ 4 ÁË 3 ˜¯

=

p 4

9.156

Chapter 9

Multiple Integrals

Example 3 Find the area of the crescent bounded by the circles r = 3 and r = 2cosq. Solution p 2 r = √3

1. The points of intersection of r = 3 and r = 2cosq are obtained as

q=

3 = 2 cos θ

P

3 cos θ = 2 π θ=± 6 Hence, θ =

p 6

q=

A

r = 2 cosq B

q q =0

O

π at P. 6

q =−

p 6

2. The region is symmetric about the Fig. 9.155 initial line, q = 0. Area of the crescent = 2 (area above the initial line, q = 0) 3. Draw an elementary radius vector OAB from the origin in the region above the initial line. OAB enters in the region from the circle r = 3 and terminates on at the circle r = 2 cosq. Limits of r : r = 3 to r = 2 cos θ π Limits of θ : θ = 0 to θ = 6 π

A = 2∫ 6 ∫

2 cos θ

r dr dθ

0

3

π

2 2 cos θ

= 2∫ 6 0

r 2

dθ 3

π

= ∫ 6 (4 cos 2 θ − 3)dθ 0

π

= ∫ 6 [2(1 + coss 2θ ) − 3] 3]dθ 0

π

6 siin n 2θ = 2 −θ 2 0

π π − 3 6 3 π = − 2 6 = sin sin

9.157

9.7 Area as Double Integral

Example 4 Find the area which lies inside the circle r = 3a cosq and outside the cardioid r = a (1 + cosq ). q=

Solution

p 2

1. The points of intersection of the circle r = 3a cos q and the cardioid r = a (1 + cos q ) are obtained as 3a cos θ = a (1 + cos cos θ ) 1 cos θ = 2 π θ=± 3

q=

p 3 r = 3a a cosq

R

B A Q

q O

P r = a (1 + cosq )

q =0

π p q =− at R. 3 3 2. The region is symmetric about the iniFig. 9.156 tial line q = 0. Total area = 2 (area above the initial line) 3. Draw an elementary radius vector OAB from the origin in the region above the initial line. OAB enters in the region from the cardioid r = a (1 + cosq ) and terminates on the circle r = 3a cosq. Limits of r : r = a (1 + cosq ) to r = 3a cosq π Limits of θ : θ = 0 to θ = 3 Hence, θ =

p

A = 2Ú 3 Ú

3 a cosq

0

a (1+ cosq )

p 3 0

r2 2

= 2Ú

r dr dq

3 a cosq

dq a (1+ cosq )

p

= Ú 3 ÈÎ9a 2 cos2 q - a 2 (1 + cos q )2 ˘˚ dq 0

p

= Ú 3 ÈÎ8a 2 cos2 q - a 2 - 2 2a a 2 cos q ˘˚ dq 0

p

= a 2 Ú 3 [ 4(1 + cos 2q ) - 1 - 2 cos q ] dq 0

p 3 4 sin 2q = a 3q + - 2 sin q 2 0 2

9.158

Chapter 9

Multiple Integrals

2p pˆ Ê p = a 2 Á 3 + 2 sin - 2 sin ˜ Ë 3 3 3¯ = p a2

Example 5 Find the area lying inside the circle r = a sin q and outside the cardioid r = a (1 – cos q). [Summer 2016] Solution 1. The points of intersection of circle r = a sinq and the cardioid r = a (1 – cosq) are obtained as a sinq = a(1 – cosq) 2 sin

q q q cos = 2 sin 2 2 2 2

sin

q q = 0, tan = 1 2 2 q q p = 0, = 2 2 4 p q = 0, q = 2

Fig. 9.157

p at P. 2 2. Draw an elementary radius vector OAB from origin in the region. OAB enters in the region from the cardioid r = a(1 – cos q) and terminates on the circle r = a sinq. Hence, q = 0 at origin and q =

Limit of r: r = a(1 – cos q) to r = a sin q Limit of q: q = 0 to q = p

A=Ú2Ú

a sin q

0

a (1- cosq )

p

r2 2

=Ú2 0

p 2

r dr dq

a sin q

dq a (1- cosq )

p

=

1 2È 2 2 a sin q - a 2 (1 - cos q )2 ˘˚ dq 2 Ú0 Î

=

a2 2

p 2 0

Ú

Èsin 2 q - (1 - 2 cos q + cos2 q )˘ dq Î ˚

9.7 Area as Double Integral

=

a2 2

p

Ú02 ÈÎsin

2

9.159

q - cos2 q + 2 cos q - 1˘˚ dq

p p p ˘ a2 È 2 Í Ú (- cos 2q )dq + 2 Ú 2 cos q dq - Ú 2 dq ˙ 0 0 ˚ 2 Î 0 p È p p ˘ a 2 Í sin 2q 2 2 - q 2˙ = + 2 sin q 0 0 ˙ 2 ÍÎ 2 0 ˚

=

=

a2 2

p p˘ È 1 ÍÎ- 2 (sin p - sin q ) + 2(sin 2 - sin 0) - 2 ˙˚

a2 2

È p˘ Í2 - 2 ˙ Î ˚ p Ê ˆ = a2 Á1 - ˜ Ë 4¯

=

Example 6 Find the area common to the cardioids r = a(1 + cosq ) and r = a(1 – cos q ). Solution 1. The points of intersection of the cardioids r = a( a (1 + cos θ ) and r = a (1 − cos cos θ ) are obtained as a (1 + cosθ ) = a( a (1 − cosθ ) cosθ = 0 π θ=± 2 Hence, θ =

q= r = a (1 − cosq )

p 2 r = a (1 + cosq )

P A

O

q =0

π at P. 2

2. The region is symmetric in all the quadrants Total area = 4 (area in the first Fig. 9.158 quadrant) 3. Draw an elementary radius vector OA from the origin in the region which lies in the first quadrant. OA starts from the origin and terminates on the cardioid r = a(1 – cos q ).

9.160

Chapter 9

Multiple Integrals

Limits of r : r = 0 to r = a (1 − coss θ ) Limits of θ : θ = 0 to θ =

π 2

p

a (1- cosq )

0

0

p 2 0

r2 2

A = 4Ú 2 Ú = 4Ú

r dr dq

a (1- cosq )

dq 0

p

= 2 Ú 2 a 2 (1 - cos q )2 dq 0

p

= 2 a 2 Ú 2 (1 - 2 cos q + cos2 q )dq 0

p

1 + cos 2q ˆ Ê = 2 a 2 Ú 2 Á 1 - 2 cos q + ˜¯ dq 0 Ë 2 3 sin 2θ = 2a θ − 2 sin θ + 2 4 2

π 2 0

 3π  = 2a 2  − 2  4 

Example 7 Find the area inside the cardioid r = 3(1 + cos θ ) and outside the 3 parabola r = . 1 + cos θ Solution 1. The points of intersection of the cardioid r = 3(1 + cos θ ) and the 3 are obparabola r = 1 + cos θ tained as

P

3 3(1 + cos cos θ ) = (1 + cos θ )

O

(1 + cos θ ) 2 = 1 cos θ = 0

π θ=± 2 π Hence, θ = at P. 2

q=

p 2

B A r = 3 (1 + cosq ) q =0

r=

3 1 + cosq

Fig. 9.159

9.7 Area as Double Integral

9.161

2. The region is symmetric about the initial line q = 0. Total area = 2 (area above the initial line) 3. Draw an elementary radius vector OAB from the origin in the region above the 3 initial line q = 0. OAB enters in the region from the parabola r = and 1 + c os θ terminates on the cardioid r = 3(1 + cos θ ). 3 Limits of r : r = to r = 3(1 + ccos q ) 1 + cos q Limits of θ : θ = 0 p

A = 2Ú 2 Ú 0

p 2 0

= 2Ú

3(1+ cosq ) r 3 1+ cosq

r2 2

to θ =

π 2

dr dq

3(1+ cosq )

dq 3 1+ cosq

p È ˘ 1 = Ú 2 9 Í(1 + cos q )2 ˙ dq 2 0 (1 + cos q ) ˚ Î p

1 + cos 2q 1 È ˘ = 9 Ú 2 Í1 + 2 cos q + ˙ dq 2 0 2 Ê 2 qˆ ˙ Í ÁË 2 cos 2 ˜¯ ˙ ÍÎ ˚ p È3 cos 2q 1 Ê qˆ q˘ = 9 Ú 2 Í + 2 cos q + - Á 1 + tan 2 ˜ sec 2 ˙ dq 0 2 2 4Ë 2¯ 2˚ Î

p È3 cos 2q 1 q 1 q Ê1 q ˆ˘ = 9 Ú 2 Í + 2 cos q + - secc 2 - ◊ tan 2 Á sec 2 ˜ ˙ dq 0 2 2 4 2 2 2Ë2 2¯˚ Î

q tan 3q sin 2q 1 q 1 2 =9 + 2 sin q + - ◊ 2 tan 2 4 4 2 2 3 3

p 2

0

È [ f (q )]n+1 ˘˙ Í∵ Ú [ f (q )]n f ¢(q )dq = n +1 ˙ Í Î ˚ p sin p 1 p 1 3 pˆ Ê 3p = 9Á + 2 sin + - tan n - tan ˜ Ë 4 2 4 2 4 6 4¯ Ê 3p 4 ˆ = 9Á + Ë 4 3 ˜¯

9.162

Chapter 9

Multiple Integrals

Example 8 Find the area common to both the circles r = cos q and r = sin q. [Winter 2013] Solution q=

1. So the point of intersection of the circles r = cos q and r = sin q is obtained as

p 2 r = sin q

cos q = sin q

P

q=

p 4

B

tan q = 1 p q= 4

A O

r = cos q q =0

p Hence, q = is the point of intersection 4 2. The point of intersection divides the region into two subregions OAP and OBP. Fig. 9.160 3. Draw an elementary radius in each subregion. (i) In subregion OAP, radius vector OA starts from the origin and terminates on the circle r = sin q. Limit of r: r = 0 to r = sin q p Limit of q: q = 0 to q = 4 (ii) In the subregion OBP, the radius vector OB starts from the origin and terminates on the circle r = cos q. Limit of r: r = 0 to r = cos q p p Limit of q: q = to q = 4 2 A=

p 4

Ú Ú 0

=

Ú

p 4

0

=

1È Í 2Í Î

1È = Í 2Í Î

sin q

0

r2 2

Ú

p 4

0

Ú

p 4

0

r dr dq +

sin q

dq + 0

Ú

p 2

Ú Ú 0

p 2 p 4

sin 2 q dq +

Ú

r dr dq

0

r2 2 p 2 p 4

cosq

cosq

dq 0

˘ cos2 q dq ˙ ˙ ˚

Ê 1 - cos 2q ˆ ÁË ˜¯ dq + 2

Ú

p 2 p 4

Ê 1 + cos 2q ˆ ˘ ÁË ˜¯ dq ˙ 2 ˙ ˚

9.7 Area as Double Integral

1È = Í 4Í Î

Ú

p 4

0

(1 - cos 2q ) dq +

È 1Í sin 2q = Íq 4 2 ÍÎ

p 4 0

Ú

p 2 p 4

9.163

˘ (1 + cos 2q ) dq ˙ ˙ ˚

sin 2q +q+ 2

=

1 ÈÊ p 1 ˆ Ê p p 1 ˆ ˘ + - - ˙ Í 4 ÎÁË 4 2 ˜¯ ÁË 2 4 2 ˜¯ ˚

=

1Êp ˆ -1 4 ÁË 2 ˜¯

=

p 1 8 4

˘ ˙ p ˙ 4˙ ˚ p 2

Example 9 Find the area common to the circles r = cos θ and r = 3 sin θ . Solution 1. The point of intersection of the circles r = cos θ and r = 3 sin θ is obtained as 3 sin n θ = ccos os θ

q=

1 tan θ = 3 π θ= 6

q=

π Hence, θ = at P. 6

B

2. Divide the region OAPBO into two subregions OAP and OBP. Draw an elementary radius vector in each subregion. (i) In subregion OAP, radius vector OA starts from the origin and terminates on the circle r = 3 sin θ. Limits of r : r = 0 to r = 3 sin θ Limits of θ : θ = 0 to θ =

p 2

π 6

p 6

P r = √3 sin q A

O

r = cos q q =0

Fig. 9.161

9.164

Chapter 9

Multiple Integrals

(ii) In the subregion OBP, the radius vector OB starts from the origin and terminates on the circle r = cos q. Limits of r : r = 0 to r = cos q Limits of q : q =

p 6

to q =

p 2

p 3 sin q 6 r 0 0

A=Ú

p 6 0

=Ú

Ú

r2 2

dr dq + Ú

3 sinn q

dq + Ú 0

Ú

r2 2

dr dq

cosq

dq 0

p

p

=

p 2 p 6

p cosq 2 r p 0 6

1 6 1 3 sin 2 q dq + Ú p2 cos2 q dq 2 Ú0 2 6

=

p

p 6 0

3 Ê 1 - cos 2q ˆ 1 2 Ê 1 + ccos os 2q ˆ ˜¯ dq + 2 Ú p ÁË ˜¯ dq 2 Ú ÁË 2 2 6

3 sin n 2q = q4 2

p 6 0

1 sin 2q + q+ 4 2

p 2 p 6

=

3Êp 1 pˆ 1Êp p 1 1 pˆ - sin ˜ + Á - + sin p - sin ˜ 4 ÁË 6 2 3¯ 4 Ë 2 6 2 2 3¯

=

5p 3 24 4

Example 10 Find the area common to the circle r = a and the cardioid r = a(1 + cos q ). Solution 1. The points of intersection of the circle r = a and the cardioid r = a (1 + cos q ) are obtained as a = a(1 + cos q ) cos q = 0 p q=± 2 π Hence, θ = at Q. 2

9.7 Area as Double Integral

9.165

2. The region is symmetric about the initial line q = 0. Total area = 2 (area above the initial line) 3. Divide the region OPQO above the initial line into two subregions OPQ and OBQ. Draw an elementary radius vector in each subregion. (i) In the subregion OPQ the radius vector OA starts from the origin and terminates on the circle r = a. Limits of r r q

p 2

Q A

B r

r

a

(1 + cos q )

O

P

q =0

Fig. 9.162

Limits of

2 (ii) In the subregion OBQ, radius vector OB starts from the origin and terminates on the cardioid r = a (1 + cos q ). Limits of + cos ) Limits of

2 Ê

p

1+ osq )

dr d

2

Ê

p

a (1+ cosq )

a

0 p

p 2

2

1+ o

ˆ

ˆ

9.166

Chapter 9

Multiple Integrals

p

1+

p 2

p a2 2

a2 p 2 5p 4

2

s 2q 4

3 2

2

2q

dq

2 p p 2

p 2

in

p a2 + (sin 2 2 4

2

Example 11 Find the area between the curve r = a (secq + cosq ) and its asymptote r = a sec q.

q = p 2

r = a (sec q + cos q )

Solution 1. The region is symmetric about the initial line q = 0 Total area = 2(area above the initial line) 2. Draw an elementary radius vector OAB in the region above the initial line. OAB enters in the region from the line r = a sec q and terminates on the curve r = a (secq + cosq ). Limits of r : r = a sec q to r = a (sec q + cos q ) Limits of q : q = 0 p

to

r dr dq

sec q

2Ú

p 2

r2 2

sec q

2

a (sec +

A

B r=

a (sec +co

dq a sec q

p

p

2

ec q q

A O

q =0

Fig. 9.163

9.7 Area as Double Integral

p ˘ È1 p = a 2 Í ◊ + 2q 02 ˙ ÍÎ 2 2 ˙˚ È p 2p ˘ = a2 Í + ˙ Î4 2 ˚

=

9.167

[ Using reduction fformula ormula ]

5p 2 a 4

Example 12 Find the area of the loop of the curve x 4 + y 4 = 8 xy. Solution 1. The equation of the curve in polar form is r 4 (cos 4 θ + si sinn 4 θ ) = 8r 2 coss θ sin sin θ r2 =

8 coss θ sin sin θ 4 coss θ + sin 4 θ

2. Draw an elementary radius vector OA from the origin in the region which lies in the first quadrant. OA starts from the origin and terminates 8 coss θ sin sin θ on the curve r 2 = . 4 coss θ + sin 4 θ Limits of 8 coss θ sin sin θ r : r = 0 to r = 4 coss θ + ssin in 4 θ π Limits of θ : θ = 0 to θ = 2 8 cossq sin q p cos4 q + sin 4 q 2 0 0

A=Ú

Ú

r dr dq

8 cossq sin q p 2 0

=Ú

2

r 2

coss4 q + sin 4 q

dq

0

p

=

1 2 Ê 8 coss q sin q ˆ dq 2 Ú0 ÁË coss4 q + sin 4 q ˜¯

π  tan n θ sec sec 2 θ  = 4∫ 2  dθ 0  1 + tan 4 θ  

q=

p 2

q =

A

r2= O

Fig. 9.164

p 4

8 cos q sin q cos4 q + sin4 q

q =0

9.168

Chapter 9

Multiple Integrals

Putting tan n 2 θ = t , 2 tan tan θ ssec ec 2 θ dθ = dt When

θ = 0, t = 0

When

θ=

π ,t → ∞ 2

A = 2Ú

•

0

dt 1 + t2

= 2 tan -1 t

• 0

Êpˆ = 2Á ˜ Ë 2¯ =p

EXERCISE 9.9 1. Find the area common to the circles r = a and r = 2acos q.   3 2 2π −  Ans. : a   2    3  2. Find the area of the crescent bounded by the circles r = 2 and r = 2 cos θ. Ans. :1 3. Find the area which lies inside the cardioid r = 2a (1 + cos q ) and 2a outside the parabola r = . 1 + cos θ  16a 2  2 Ans. : 3 π a +   3   4. Find the area bounded between the circles r = 2a sin q, r = 2b sin q (b > a).  Ans. : π(b 2 − a 2 ) 5. Find the area outside the circle r = a and inside the cardioid r = a(1 + cos q ).   a2  Ans. : (π + 8) 4  

9.8

9.8

Volume as Triple Integral

9.169

VOLUME AS TRIPLE INTEGRAL

Volume, V of a solid contained in the region V is given as V = ÚÚÚ dx dy dz V

In cylindrical coordinates, V = ÚÚÚ r dr dq dz V

In spherical coordinates, V = ÚÚÚ r 2 sinq q dr dq df V

Note: Consider symmetricity of the region while calculating the volume.

Example 1 Find the volume of the tetrahedron bounded by the plane x + y + z = 2 and the planes x = 0, y = 0, z = 0. [Winter 2014] Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from xy-plane and terminates on the plane x + y + z = 2. Limits of z: z = 0 to z = 2 – x – y 2. Projection of the plane x + y + z = 2 in xy-plane is the DOPQ. Equation of the line PQ is obtained by putting z = 0 in x + y + z = 2. 3. Draw a vertical strip A¢B¢ in the region OPQ in the xy-plane. A¢B¢ starts from the x-axis and terminates on the line x + y = 2.

Fig. 9.165

Limits of y : y = 0 Limits of x : x = 0

to to

y=2–x x=2

9.170

Chapter 9

Multiple Integrals

V =Ú

2 2- x

=Ú

2 2- x

=Ú

2 2- x

=Ú

2

2- x - y

Ú Ú0

0 0

Ú

0 0

dz dy dx

2- x - y

z0

dy dx

Ú [(2 - x) - y ] dy dx

0 0

0

y2 (2 - x ) y 2

2- x

dx 0

2È ( 2 - x )2 ˘ = Ú Í ( 2 - x )2 ˙ dx 0 2 ˙˚ ÍÎ

=Ú

2

0

( 2 - x )2 dx 2

1 (2 - x )3 = 2 -3

2

0

2 1 = - (2 - x )3 0 6 1È = - Î0 - (2)3 ˘˚ 6 4 = 3

Example 2 Find the volume of the tetrahedron bounded by the planes x = 0, y = 0, x y z z = 0 and + + = 1. a b c Solution 1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from x y z xy-plane and terminates on the plane + + = 1. a b c x yˆ Ê Limits of z : z = 0 to z = c Á 1 - - ˜ . Ë a b¯ Ê x yˆ c Á 1- - ˜ Ë a b¯ 0

V =Ú

dz dy dx

9.8

2. Projection of the plane

x y z + + a b c

Volume as Triple Integral

9.171

1 in xy-plane is the ∆OPQ Equation of the line

PQ is obtained by putting z = 0 in

x y z + + a b c

x a

b

.

3. Draw a vertical strip A¢B¢ in the region OPQ. DOPQ starts from the x-axis and x y terminates on the line + = 1. a b x a

Limits of y y Limits of x x z

R (0, 0, c) y (0, b)

y x z + + b a c

1 B

B

y x a + b (0, b, 0) y

A

O

P (a, 0, 0) x

Fig. 9.166 a

b 1-

x

x y c 1- -

y x

0 a

b 1

a

b

Ú Ú =

a

x a

x y c 1- a b

x

x y a

x

x a y2 2b

y b b 1

x a

dx 0

A

P (a, 0) x

1

9.172

Chapter 9

Multiple Integrals

a

=

bc = 2 abc = 6

-

2

x a

-

x 1a 3

-

3

b2 x 12b a

2˘

x

a

0

Example 3 z2

Find the volume cut off from the sphere z2

a 2 by the cone

Solution 1. The region is bounded by the cone and sphere. Putting x = r cos q, y = r sin q, z = z, z2

(i) equation of the sphere

a 2 reduces to

z 2 reduces to

(ii) equation of the cone

2

a2 . 2

,

.

2. Draw an elementary volume AB parallel to z-axis in the region. AB starts from the cone r = z and terminates on the sphere a2 . a2

Limits of a2 r 2

V

r

r2 dq

r

z

B

r + z2 r

a2

z

p 2 A r

A O

q

y

x

Fig. 9.167

a √2 q 0

9.8

Volume as Triple Integral

9.173

3. Projection of the region in rq -plane is the curve obtained by the intersection of the sphere r 2 + z 2 = a 2 and the cone r = z as r 2 + r 2 = a2 r=

a 2

which is a circle with centre at the origin and radius

a 2

.

4. The region (circle) is symmetric in all the quadrants. Draw an elementary radius vector OA¢ in the first quadrant of the region. OA¢ starts from the origin and termia . nates on the circle r = 2 a Limits of r : r = 0 to r = 2 π Limits of θ : θ = 0 to θ = (first quadrant) 2 p

a

V = 4Ú 2 Ú 0

0

p

= 4Ú 2 Ú 0

p

2

r zr

a

È ÍÎr

2

0

p

= 4Ú 2 dq Ú 0

a2 - r 2

a

0

= 4Ú 2 Ú 0

Úr

2

(

a 2

0

r dz dr dq

a2 - r 2

dr dq

)

˘ a 2 - r 2 - r ˙ dr ˚

È 1 2 2( 2˘ Í - 2 a - r -2r ) - r ˙ dr Î ˚ a

p 2 0

3 2(a 2 - r 2 ) 2

2

È ˘ [ f (r )]n +1 n , n π -1˙ ¢(r )dr = Í∵ Ú [ f (r )] f ¢( 3 n +1 0 ÍÎ ˙˚ 3 È ˘ Ï ¸ 3 ˙ Ô p Í 1 ÔÊ 2 a 2 ˆ 2 a = 4 ◊ Í - ÌÁ a - ˜ - a 3 ˝ ˙ 2 Í 3 ÔË 2¯ Ô 6 2˙ Ó ˛ Î ˚ 3 3 È a a ˘ = 2p Í + ˙ ÍÎ 3 2 3 ˙˚ =4q

=

1 - ◊ 2

p a3 (- 2 + 2) 3

r3 3

9.174

Chapter 9

Multiple Integrals

Example 4 Find the volume bounded by the cone x2 + y2 = z2 and the paraboloid x2 + y2 = z. Solution 1. The region is bounded by the cone and the paraboloid. Putting x = r cos q, y = r sin q, z = z, (i) equation of the cone x2 + y2 = z2 reduces to r2 = z2, r = z. (ii) equation of the paraboloid x2 + y2 = z reduces to r2 = z. 2. Draw an elementary volume AB in the region. AB starts from the paraboloid r2 = z and terminates on the cone r = z. 2

Limits of z z

r r

V

r

dq

r

z

r=z

= p 2

B r2 = z A′ A O

r=1 y

O

=0

x

Fig. 9.168

3. Projection of the region in rq -plane is the curve obtained by the intersection of the cone r = z and the paraboloid r 2 = z as r = r 2, r = 1, which is a circle with centre at the origin and radius 1. 4. The region (circle) is symmetric in all the quadrants. Draw an elementary radius vector OA¢ in the first quadrant of the region. OA¢ starts from the origin and terminates on the circle r = 1. Limits of r r Limits of q q

p (first quadrant) 2

9.8

p

1 r

V

0 r p

dq

r

1

9.175

Volume as Triple Integral

r

4Ú 2 Ú r z r 2 dr dq p

1

p 0

.

r3 3

p 1 2 3

r4 4

1

0

1 4

p 6

Example 5 Find the volume of the solid bounded by the paraboloids and y

y

Solution 1. The region is bounded by the paraboloids. Putting

si

z

z,

2

(i) equation of the paraboloid

y reduces to z = r .

(ii) equation of the paraboloid

y

reduces to z = 4 – 3r2.

2. Draw an elementary volume AB in the region. AB starts from the paraboloid z = r2 and terminates on the paraboloid z = 4 – 3r2. 2 r2 Limits of 4 - 3r 2

V

dq

r

z

q= p 2 B

A O

3r 2

z

4

z

r2 y

A r O

x

Fig. 9.169

1 =0

9.176

Chapter 9

Multiple Integrals

3. Projection of the region in all the quadrants is the curve obtained by the intersection of the paraboloids z = r2 and z = 4 – 3r2 as r 2 = 4 - 3r 2 r2 = 1 r=1 which is a circle with centre at the origin and radius 1. 4. The region (circle) is symmetric in all the quadrants. Draw an elementary radius vector OA¢ in the first quadrant of the region. OA¢ starts from the origin and terminates on the circle r = 1. Limits of r : r = 0 to r = 1 Limits of q : q = 0 to q =

p 2 p

V = 4Ú 2 Ú

1 4 - 3r 2

Ú

r 0 0 r2 p 1 4 - 3r 2 2 r z r2 0 0

= 4Ú

Ú

p

dz dr dq dr dq

1

= 4Ú 2 Ú r (4 - 3r 2 - r 2 )dr dq 0

0

p

1

= 4Ú 2 dq Ú (4r - 4r 3 )dr 0

0

p

= 4 q 02 2 2rr 2 - r 4 = 4◊

1 0

p 2

= 2p

Example 6 Find the volume of the cylinder x2 + y2 = 2ax intercepted between the x 2 + y2 paraboloid z = and the xy-plane. 2a Solution 1. The region is bounded by the cylinder, paraboloid and xy-plane. Putting x = r cos q, y = r sin q, z = z (i) equation of the cylinder x2 + y2 = 2ax reduces to r2 = 2ar cos q, r = 2a cos q. (ii) equation of the paraboloid z =

r2 x 2 + y2 reduces to z = . 2a 2a

9.8

Volume as Triple Integral

9.177

2. Draw an elementary volume AB parallel to z-axis in the region. AB starts from r2 xy-plane and terminates on the paraboloid z 2a r2 2a

Limits of z : z = 0 to z

r2

V

dq

r

3. Projection of the region in rq -plane is the curve obtained by the intersection of the cylinder r = 2a cos q and rq -plane (z = 0) as r = 2a cos q, which is circle with centre (a, 0) and radius a. 4. The region (circle) is symmetric about the initial line q = 0. Draw an elementary radius vector OA¢ in the region above initial line q = 0. OA¢ starts from the origin and terminates on the circle r = 2a cos q. Limits of r : r = 0 to r = 2a cos q Limits of q : q = 0 to

(first quadrant)

2

z

q= p 2

r2 z= 2a

A

B

r = 2 cos q y r = 2a cos q

x

Fig. 9.170 p

r2

2 a co q

V

r

0 p 2 0

Ú

2 a cosq

2Ú 2 Ú

2 a cosq

2Ú

p

0

p

1 2 r aÚ 4

2a

rz

r2 2 a dr dq 0

r2 ˆ dr dq 2a

r sq

dq 0

dq

O

=0

9.178

Chapter 9

Multiple Integrals

p

= 4 a 3 Ú 2 cos4 q dq 0

3 1 p = 4a3 ◊ ◊ ◊ 4 2 2 3 3p =a ◊ 4 3 = p a3 4

[ Using reduction formula ]

Example 7 Find the volume bounded by the cylinder x2 + y2 = 4 and the planes y + z = 3, z = 0. [Summer 2014] Solution 1. The region is bounded by the cylinder x2 + y2 = 4 and the planes y + z = 3, z = 0. 2. Draw an elementary volume AB parallel to z-axis in the region. AB starts on xy-plane and terminates on the plane y + z = 3. Limits of z: z = 0 to z = 3 – y 3. Projection of the region in xy-plane is the circle x2 + y2 = 4.

Fig. 9.171

4. The region in xy-plane is symmetric in all the quadrants. 5. Draw a horizontal strip A¢B¢ in the first quadrant of the circle. A¢B¢ starts from y-axis and terminates on the circle x2 + y2 = 4. Limits of x : x = 0 to x = Limits of y : y = 0 to y = 2

4 - y2

9.8

V = 4Ú

2

= 4Ú

2

= 4Ú

2

Ú

4 - y2

0 0

Ú

4 - y2

0 0

Ú

4 - y2

0 0 2

3- y

Ú0

dx dy

(3 - y) dx dy

= 4Ú (3 - y) x 0 0

9.179

dz dx dy

3- y

z0

Volume as Triple Integral

4 - y2

dy

2

= 4Ú (3 - y) 4 - y 2 dy 0

2 2 = 4 ÈÍ Ú 3 4 - y 2 dy - Ú y 4 - y 2 dy ˘˙ 0 Î 0 ˚

È 3 2 y 4 y 1 = 4 Í3 4 - y 2 + sin -1 + (4 - y 2 ) 2 Í 2 2 20 3 ÍÎ 3˘ È 1 ˆ2˙ Ê Í -1 = 4 Í3 Á 2 sin 1 + (4)˜ ˙ Ë 3 ¯ ˙˚ ÎÍ

2˘

˙ ˙ 0˙ ˚

8˘ È = 4 Í3p + ˙ 3˚ Î 4 = [ 9p + 8] 3

Example 8 Find the volume bounded by the cylinders x2 + y2 = 2ax and z2 = 2ax. Solution 1. The region is bounded by the circular and parabolic cylinders. Putting x = r cos q, y = r sin q, z = z, (i) equation of the cylinder x2 + y2 = 2ax reduces to r2 = 2ar cos q, r = 2a cos q. (ii) equation of the cylinder z2 = 2ax reduces to z2 = 2ar cos q. 2. The region is bounded by the cylinder z2 = 2ar cos q. arr cos cos q Limits of z : z = - 2 ar cos q to z = 2 a V = ÚÚ Ú

2 ar cosq - 2 ar cosq

r dz dr dq

3. Projection of the region in rq -plane is the circle r = 2a cos q.

9.180

Chapter 9

Multiple Integrals

4. The region in rq -plane (circle) is symmetric about the initial line q = 0. Draw an elementary radius vector OA in the region above the initial line q = 0. OA starts from the origin and terminates on the circle p q = r = 2a cos q. 2 Limits of r : r = 0 to r = 2a cosq p Limits of q : q = 0 to q = (first 2 quadrant) A

p 2 a cosq 2 ar cosq 2 0 0 - 2 ar cosq

V = 2Ú

Ú

Ú

p 2 a cosq 2 0 0

= 2Ú

Ú

p

= 2Ú 2 Ú 0

p 2 0

= 4Ú

2 a co ossq

0

z

2 ar cosq - 2ar 2 cosq

O

q =0

r dr dq 1

2(2ar 2 ar cos q ) 2 r dr dq

1 1 (2 a ) 2 (cos q ) 2

p 2 0

r = 2a cos q

r dz dr dq

5 2 2r

5

Fig. 9.172

2 a cosq

0

1 1 5 (2 a ) 2 (cos q ) 2 2(2 a cos cos q ) 2 dq

=

4 5Ú

=

64 a 3 5

=

64 a 3 2 ◊ 5 3

=

128a 3 15

p 2 0

Ú

cos3 q dq [ Using reduction fformula ormula ]

Example 9 Use triple integral to find the volume of the solid within the cylinder x2 + y2 = 9 between the planes z = 1 and x + z = 1. [Summer 2017, 2015] Solution 1. The region is bounded below by the plane x + z = 1 and above by the plane z = 1. Limits of z: z = 1 – x to z = 1 2. Draw an elementary volume AB parallel to z-axis in the region. AB starts from the plane x + z = 1 and terminates on the plane z = 1. 3. Projection of the region in xy-plane is the right half of the circle x2 + y2 = 9.

9.8

Volume as Triple Integral

9.181

Fig. 9.173

4. The region in xy-plane is symmetric about x-axis. Draw a horizontal strip in A¢B¢ in the region. A¢B¢ starts from y-axis and terminates on the circle x2 + y2 = 9. 9 - y2

Limits of x : x = 0

to

x=

Limits of y : y = 0

to

y = 3 (first quadrant) 9 - y2

V = 2Ú

3

= 2Ú

3

9 - y2

= 2Ú

3

9 - y2

= 2Ú

3

Ú

0 0 0 Ú0 0 Ú0

0

x2 2

1

Ú1- x dz dx dy 1

z 1- x dx dy (1 - 1 + x ) dx dy

9 - y2

dy 0

3

= Ú (9 - y 2 ) dy 0

y3 = 9y 3

3

0

= 27 - 9 = 18

Example 10 Find the volume in the first octant bounded by the circular cylinder x2 + y2 = 2 and planes z = x + y, y = x, z = 0, x = 0. Solution 1. The region is bounded below by xy-plane (i.e. z = 0) and above by the plane z = x + y.

9.182

Chapter 9

Multiple Integrals

Limits of z : z = 0 to z = x + y V = ÚÚ Ú

x+ y

0

dz dy dx

2. Projection of the region in xy-plane is the region bounded by the circle x2 + y2 = 2, the line y = x and x = 0. y 3. The point of intersection of the circle x2 + y2 = 2 and the line y = x is obtained as y=x Q

x2 + x2 = 2 2 x2 = 2 x = ±1 \ y = ±1 The point of intersection is P (1, 1). 4. Draw a vertical strip AB parallel to yaxis which starts from the line y = x and terminates on the circle x2 + y2 = 2.

B

A

P(1, 1) x2 + y2 = 2

O

x

Fig. 9.174

Limits of y : y = x to y = 2 - x 2 Limits of x : x = 0 to x = 1 V =Ú

1

=Ú

1

=Ú

1

=Ú

1

2 - x2

Ú

0 x 2 - x2

Ú

0 x 2 - x2

Ú

0 x

0

x+ y

Ú0

dz dy dx

x+ y

z0

dy dx

( x + y)dy dx

y2 xy + 2

2 - x2

dx x

1È 1 ˘ = Ú Í x 2 - x 2 - x + ( 2 - x 2 - x 2 )˙ dx 0Î 2 ˚ 1È 1 ˘ 2 = Ú Í2 - x 2 ( -2 x ) - 2 x + 1˙ dx 0Î 2 ˚

(

)

(

)

3 x2 ) 2

1 2(2 = - . 2 3 ==

(

)

1

2x +x 3 0

1 2 3 - +1 2 3 1- 2 3

2 2 3

3

n +1 È ˘ f ( x )] [ n Í∵ Ú [ f ( x )] f ¢( x )dx = , n π -1˙ n +1 Î ˚

9.8

9.183

Volume as Triple Integral

Example 11 Find the volume of the wedge intercepted between the cylinder x2 + y2 = 2ax and the planes z = mx, z = nx where n > m. Solution 1. The region is bounded below by the plane z = mx and above by the plane z = nx. Limits of z : z = mx to z = nx nx mx

y

B

y x

2

2. Projection of the region in the xy-plane is bounded by the circle x2 + y2 = 2ax. 3. The region (circle) in xy-plane is symmetric about x-axis. Draw a vertical strip AB parallel to y-axis in the region above x-axis. AB starts from the x-axis and terminates on the circle x2 + y2 = 2ax. Limits of y : y = 0

to

Limits of x : x = 0

to

2a

2a

Ú

x

A

Fig. 9.175

x = 2a

x=0 y 0

2Ú

P (2a, 0) O

2

2 ax x 2

2a

+ y = 2ax

nx

y x

z mx

2 ax - x 2

nx

x

2 ax - x 2

Ú

2a

Ú

2a

mx

x y0

2a

2a

˘ y dx

2 ax x 2

x

2 ax x 2

x

dx

x 2 dx

3

x dx

Putting x = 2a cos2 q, dx = – 4a cosq sinq dq p When x = 0, cos q = 0, q 2 When x = 2a, cos q = 1, q = 0 0 2

3

os2 ) 2

c

( 4 a cos

dq

9.184

Chapter 9

Multiple Integrals

p

6a3

co

4

in

1 3 1 p 6 4 2 2

dq Using r

nf

]

a3

Example 12 A cylindrical hole of radius b is bored through a sphere of radius a. Find the volume of the remaining solid. Solution Let the equation of the cylinder with radius b is x2 + y2 = b2 and equation of the sphere with radius a is x2 + y2 + z2 = a2. 1. The region is bounded by the sphere and cylinder. Putting x = r cos q, y = r sin q, z = z, (i) equation of the cylinder x2 + y2 = b2 reduces to r2 = b2, r = b. (ii) equation of the sphere x2 + y2 + z2 = a2 reduces to r2 + z2 = a2. 2. The region is symmetric in all the octants. Volume = 8 (volume in the positive octant) 3. Draw an elementary volume AB in the positive octant of the region. AB starts from xy-plane and terminates on the sphere r2 + z2 = a2. 2

Limits of z z a2 r 2

V

r2 dq

r

z

q r

b r 2 + z2

a B′

B O y

O

p 2

A′

r

b q =0 r

A

x

Fig. 9.176

a

9.8

Volume as Triple Integral

9.185

4. Projection of sphere r2 + z2 = a2 in rq -plane (z = 0) is r2 = a2, r = a. Projection of the region in rq -plane is the region bounded by the circles r = b and r = a. 5. Draw an elementary radius vector OA¢B¢ in the region in the first quadrant. OA¢B¢ enters in the region from the circle r = b and terminates on the circle r = a. Limits of r : r = b to r = a p 2

Limits of q : q = 0 to q = p

V = 8Ú 2 Ú

a

b Ú0

0

p

a2 - r 2

0

b

p

r dz dr dq

a2 - r 2

a

= 8Ú 2 Ú r z

(first quadrant)

dr dq

0

a

= 8 Ú 2 Ú r a 2 - r 2 d r dq 0

b

p 2 0

È a 1 = 8 Ú dq Í Ú Î b 2 = -4 q

p 2 0

2(a

2

(

˘ a 2 - r 2 ( -2r )˙ dr ˚

3 - r2 )2

3

)

a

b

3 È 2 2 2 Í p 2(a - b ) = - 4 ◊ ◊ Í0 2 Î 3

È [ f (r )]n +1 ˘ n Í∵ Ú [ f (r )] f ¢(r ) dr = ˙ n +1 ˚ Î ˘ ˙ ˙ ˚

3

=

4p 2 (a - b2 ) 2 . 3

Example 13

2

2

2

Ê xˆ 3 Ê yˆ 3 Ê zˆ 3 Find the volume bounded by the solid Á ˜ + Á ˜ + Á ˜ = 1. Ë a¯ Ë b¯ Ë c¯ Solution It is difficult to integrate this integral in Cartesian form. Therefore, we transform the solid into a sphere using following change of variables. Putting x = au3, y = bv3, z = cw3 2

2

2

Ê xˆ 3 Ê yˆ 3 Ê zˆ 3 ÁË a ˜¯ + ÁË b ˜¯ + ÁË c ˜¯ = 1 reduces to u2 + v2 + w2 = 1, which is a sphere of radius 1 and centre at the origin. and

dx dy dz = |J | du dv dw

9.186

Chapter 9

Multiple Integrals

∂x ∂u ∂( x, y, z ) ∂y J= = ∂(u, v, w) ∂u ∂z ∂u

where

3au2 =

0

0

3bv

0

0

∂x ∂v ∂y ∂v ∂z ∂v

∂x ∂w ∂y ∂w ∂z ∂w

0 2

0 3cw 2

= 27abc u2 v 2 w2 \ dx dy dz = 27 abc u 2 v 2 w2 du dv dw In the new coordinate system u, v, w, the region is bounded by a sphere. Putting u = r sin q cos f, v = r sin q sin f, w = r cos q, sphere u2 + v2 + w2 = 1 reduces to r = 1. Since region is symmetric in all the octants. Volume = 8 (volume in the positive octant) Limits in the positive octant of the sphere are Limits of r : r = 0 to r = 1 Limits of q : q = 0

to

θ=

π 2

Limits of f : f = 0

to

φ=

π 2

V = 8ÚÚ ÚÚÚÚ dx dy dz 2 2 2 = 8ÚÚÚ ÚÚÚ 27abc u v w du dv dw p

p

0

0

1

= 216 abc Ú 2 Ú 2 Ú (r 2 sin 2 q cos2 f )(r 2 sin 2 q ssiin 2 f )( )(r 2 cos cos2 q )r 2 ssin q dr dq df 0

p

p

0

0

1

= 216 abc Ú 2 coss2f sin sin 2 f df Ú 2 sin sin 5q cos2 q dq Ú r 8 dr Ê 1 1 p ˆ Ê 4 2 1ˆ r9 = 216 abc ◊ Á ◊ ◊ ˜ ◊ Á ◊ ◊ ˜ Ë 4 2 2 ¯ Ë 7 5 3¯ 9 =

4 p abc 35

0

1

[Using reduction formula] 0

9.8

Volume as Triple Integral

9.187

EXERCISE 9.10 1. Find the volume of the sphere x 2 + y 2 + z 2 = a 2 cut off by the planes z = 0 and the cylinder x2 + y2 = ax.  2 3  π 2   Ans. : 3 a  2 − 3     2. Find the volume common to the sphere x 2 + y 2 + z 2 = a 2 and the cylinder x 2 + y 2 = b 2 , (a > b).  Ans. : 2π b 2 a 2 − b 2  3. Find the volume bounded by the cylinders y 2 = x, x 2 = y and the planes z = 0, x + y + z = 2. 11    Ans. : 30    4. Find the volume of the cylinder x 2 + y 2 − 4 x = 0 cut by the cylinder z 2 = 4 x. x

1024    Ans. : 15    2 2 5. Find the volume of the paraboloid x + y = 4 z cut off by the plane z = 4.

[ Ans. : 32π ] 2 y 6. Find the volume of the paraboloid x 2 + + z = 1 cut off by the plane z = 0. 9 3π    Ans. : 2    7. Find the volume of the solid bounded by the paraboloids y2 y4 z = 4 − x2 − and z = 3x 2 + . 4 4

 Ans. : 4 2π 

8. Find the volume of the solid bounded by the plane z = 0, the paraboloid 3z = x 2 + y 2 and the cylinder x 2 + y 2 = 9. 27π    Ans. : 2    9. Find the volume of the solid bounded by the planes z = 0, z = 3 and the cylinders y = x 2 , x 2 = y . [Ans. : 1] 10. Find the volume of the solid which is above the cone z 2 + x 2 + y 2 and inside the sphere x 2 + y 2 + (z − a)2 = a 2 . [Ans. : p a3]

9.188

Chapter 9

Multiple Integrals

POINTS TO REMEMBER • Double Integrals The double integral of a function f (x, y) over the region R is denoted by

∫∫ f(x, y)dx dy. R

• Evaluation of Double Integrals

Method-I: When the region R is bounded by the curves y = y1(x), y = y2(x) and x = a, x = b,

∫∫ f(x, y)dx dy = ∫ ∫ R

b

y2 (x)

a

y1 ( x )

f (x, y )dy  dx 

Method-II: When the region R is bounded by the curves x = x1(y), x = x2(y) and y = c, y = d,

∫∫ f(x, y)dx dy = ∫ ∫ R

d

x2 ( y )

c

x1( y )

f (x, y )dx  dy 

If all the four limits are constant and f (x, y) is explicit, then the f (x, y) can be integrated w.r.t. any variable first and also can be written as product of two single integrals. • Change of Order of Integration Sometimes evaluation of double integral becomes easier by changing the order of integration. To change the order of integration, 1. Draw the region of integration with the help of the given limits. 2. Draw vertical or horizontal strip as per the required order of integration 3. Find the limits of integration

∫ ∫ b

y2 (x)

a

y1 ( x )

f (x, y )dy  dx = 

∫ ∫ d

x2 ( y )

c

x1( y )

f (x, y )dx  dy 

• Double Integrals in Polar Coordinates Putting x = r cos q, y = r sin q, sin θ) J dr dθ ∫∫ f(x, y)dy dx = ∫∫ f(r cos θ,r sin where Hence,

Jacobian, J = r sin θ)r dr dθ ∫∫ f(x, y)dy dx = ∫∫ f(r cos θ,r sin (Contd )

Points to Remember

9.189

(Contd ) • Triple Integrals The triple integral of a continuous function f (x, y, z) over a region V is denoted by ÚÚÚÚ f (x, y , z)dx dy dz. V

• Triple Integrals in Cartesian Coordinates If the region V is bounded below by a surface z = z1 (x, y) and above by a surface z = z2 (x, y) and if the projection of region V in xy-plane is R which is bounded by the curves y = y1 (x), y = y2 (x) and x = a, x = b then b



∫∫∫ f(x, y, z)dx dy dz = ∫ ∫ a

V

y2 (x) y1 ( x )

{∫

z2 ( x , y ) z1( x , y )

}

ff((x, y, y , z)dz dy  dx 

• Triple Integrals in Cylindrical Coordinates Putting x = r cos q, y = r sin q, z = z, os q , r sin q , z) J d dzz dr dq ÚÚÚ f(x, y, z)dx dy dz = ÚÚÚ f(r ccos where

Jacobian, J = r

Hence,

os q , r sin q , z)r d dzz dr dq ÚÚÚ f(x, y, z)dx dy dz = ÚÚÚ f(r ccos

• Triple Integrals in Spherical Coordinates Putting x = r sin q cos f, y = r sin q sin f, z = r cos q, coss φ, r sin sin θ sin sin φ, r cos θ) J dr dθ dφ ∫∫∫ f(x, y, z)dx dy dz = ∫∫∫ f(r sinθ co where Hence,

Jacobian,J = r 2 sin q

∫∫∫∫ f(x, y, z)dx dy dz = ∫∫∫ f (r sin θ cos cos φ, r sin θ sin φ, r cos θ) ⋅ r

2

sin q dr dq df

Area as Double Integral • Area in Cartesian Coordinates (i) The area bounded by the curves y = y1(x) and y = y2(x) intersecting at the points P (a, b) and Q (c, d ) is A=

c

y2 (x)

a

y1 ( x )

ÚÚ

dy dx (Contd )

9.190

Chapter 9

Multiple Integrals

(Contd ) (ii) The area bounded by the curves x = x1(y) and x = x2(y) and intersecting at the points P (a, b) and Q (c, d ) is c

y2 (x)

a

y1 ( x )

ÚÚ

A=

dy dx

• Area in Polar Coordinates The area bounded by the curves r = r1(q ), r = r2(q ) and the lines q = q1 and q = q2 is q2

r2 (q )

q1

r1(q )

Ú Ú

A=

r dr dq

• Volume in Cartesian Coordinates V = ÚÚÚ dx dy dz V

• Volume in Cylindrical Coordinates V =

ÚÚÚ r dr dq dz V

• Volume in Spherical Coordinates V =

ÚÚÚ r

2

sinq dr dq df

V

MULTIPLE CHOICE QUESTIONS Choose the correct alternative in the following questions: 1. To evaluate

•

x

Ú0 Ú0 xe

-

x2 y

dx dy by change of order of integration, the lower limit

for the variable x is equal to (b) 0 (a) y2 2.

(c) •

(d) y

4 3 2

Ú0 Ú0 Ú0 dx dy dz = (a) 9

(b) 24

(c) 1

(d) 0 2 ex

3. By changing the order of integration, the integral the double integral _______. e 2

(a) (c)

Ú1 Úlog y dx dy 1

log y

Úe Ú2 2

dx dy

(b) (d)

e2

2

e2

log y

Ú0 Ú1

Ú1 Úlog y dx dy Ú1 Ú2

dx dy

dy dx is equivalent to

Multiple Choice Questions

9.191

ÚÚÚ dy dx dz, where R is the region

4. By changing to spherical polar co-ordinates,

R

of hemisphere x 2 + y 2 + z 2 = a 2 is equivalent to triple integral ______ p

2p

5.

a 2

(a)

Ú0 Ú02 Ú0 r

(c)

p p a 2 2 2 r 0 0 0

a

Ú Ú Ú x

p

(b)

sin q dr dq df

(d)

Ú Ú Ú

(c)

a4 48

y

Ú0 Ú0 Ú0 xyz dz dy dx a6 24

(a)

2p

a 2

Ú0 Ú0 Ú0 r

sin q dr dq df

p p a 2 2 2 r 0 0 0

sin q dr dq df cos q dr dq df

= a6 48

(b)

(d)

a4 24

6. In evaluating Ú Ú xy ( x + y) dx dy over the region between y = x2 and y = x, the limits are (a) x = 0 to 1, xy = 0 to 1 (b) x = 0 to 1, y = 0 to x (c) x = 0 to 1, y = 0 to x2 (d) x = 0 to 1, y = x2 to x 7.

a2 - y2

a

Ú0 Ú0

( a 2 - x 2 - y 2 ) dx dy =

p a4 8

(a)

(b)

p a4 4

(c)

8. After transforming to polar co-ordinates p 1 -r 2 2 e dr 0 0

(a)

Ú Ú

(c)

Ú02 Ú0 e

p

9.

p

a cosq

Ú0 Ú0 (a)

10.

1 2

p

11.

5 3

2

(d) pa4

• • - ( x 2 + y2 )

Ú0 Ú0 e

p 1 -r 2 2 e r 0 0

dq

(b)

Ú Ú

r dr dq

(d)

Ú02 Ú0 e

p

• - y2

dx dy =

d r dq

r dr dq

r sin q dr dq =

a2 4

Ú0 Ú0 xy (a)

• -r 2

p a4 2

(b)

a2 3

(c)

a2 2

(d)

a2 6

(b)

1 3

(c)

2 3

(d)

4 3

(c)

p 2

(d) 2

dy dx =

p

Ú02 Ú02 sin ( x + y) dx dy is (a) 0

(b) p

9.192

Chapter 9

Multiple Integrals

12. The value of the integral Ú Ú xy dx dy over the region bounded by the x-axis, ordinate at x = 2a and the parabola x2 = 4ay is a4 3

(a)

(b)

13. The triple integral

a4 5

(c)

ÚÚÚ dx dy dz

a4 7

(d)

a4 9

gives

R

(a) volume

(b) area 1 1

14. The value of 4 35

(a)

1- x

Ú0 Úy Ú0 2

(b)

p 3

(b)

3 35

p 2

(b)

(c) 2p

p

1 2

Ú0 Ú03 Ú0 r

p 6

16. The value of the integral (a)

(d)

6 35

(d)

p 4

(d)

p 6

sin q dr dq df is (c)

• • - x 2 (1 + y2 )

Ú0 Ú0 e

p 3

8 35

(c)

2p 3 x dx dy is p 4

17. Changing the order of integration in the double integral to

(d) density

x dz dx dy is

15. The value of the integral (a)

(c) surface area

s q

• 2

Ú0 Ú x f ( x, y) dy dx

leads

4

Úr Ú p f ( x, y) dx dy , then q is (b) 16y2

(a) 4y

(c) x (d) 8 2 2 ( x + y ) d x d y 18. The limits of integration of Ú Ú over the domain bounded by y = x2 and y2 = x are (a) x = 0 to 1, y = x2 to x (b) x = 0 to 1, y = 0 to 1 2 (c) x = y to y , y = 0 to 1 (d) x = 0 to y, y = x to x2 19.

ÚÚ

xy 1- y

ÚÚr

dx dy over the positive quadrant of the circle x2 + y2 = 1 is

1 6

(a) 20.

2

3

(b)

2 3

(c)

5 6

(d)

dr dq over the region included between the circles r = 2 sinq and

r = 4 sinq is (a)

5 3

p

4sin q 3 r 2sin q

Ú0 Ú

dr dq

(b)

p 4sin q 3 2 r 0 2sin q

Ú Ú

dr dq

9.193

Multiple Choice Questions

(c)

p

4sin q 3 r - p 2sin q

Ú Ú

21. On converting into polar co-ordinates p 2 r2 0 0 p a 4 r3 0 0 a

p 4sin q 3 2 r 0 sin q

Ú Ú

(d)

dr dq

a

Ú0 Ú0

a2 - x 2

a

(a)

Ú Ú

dr d q

(b)

(c)

Ú Ú

dr dq

(d)

( x 2 + y 2 ) dy dx =

p 2 r3 0

dr dq

p 4 r2 0

dr d q

Ú0 Ú a

dr dq

Ú0 Ú

22. In spherical co-ordinates, dx dy dz is equal to (a) r dq df dr (d) r sinq dq df dr (c) r2 sinq dq df dr (d) r2 dq df dr 23. The value of the integral (a) 1 a

Ú0

dx Ú

2

+ y 2 + z 2 ) dz dy dx is

1 3

(b)

24. The value of

1 1 1

Ú0 Ú0 Ú0 ( x

a2 - x 2

0

2 3

(c) dy Ú

a2 - x 2 - y2

0

(d) 3

dz is

p 2 p a3 a (c) 4pa3 (d) 3 6 25. The transformations x + y = u, y = uv transform the area element dy dx into J du dv, where J is equal to (a) 4pa2

(b)

(a) 1

(b) u

26. The value of

(c) –1

(d) u2

3 ÚÚ x y dx dy, where R is region enclosed by the ellipse R

x2 a2

+

y2 b2

in the first quadrant is (a)

b2 a 4 24

(b)

b3 a 4 24

28.

0

y

Ú1 Ú1

(c)

Ú0 Ú1

2p

Ú0

1 y

1 x

ba 4 24

(d)

Ú0 Ú0

f ( x, y)dy dx =

f ( x, y)dx dy

(b)

Ú1 Úy f ( x, y)dx dy

f ( x, y)dx dy

(d)

Ú0 Úy f ( x, y)dx dy

27. By changing the order of integration, (a)

(c)

b2 a 2 24

0 1

1 1

1

dq Ú e2r dr is equal to 0

(a) e2 – 1

(b)

p 2 (e - 1) 2

(c) p(e2 – 1)

(d) 2p(e2 – 1)

=1

9.194

Chapter 9

Multiple Integrals

ÚÚ x

29. The value of

2 3

y dx dy, where R is the region bounded by the rectangle

R

0 £ x £ 1 and 0 £ y £ 3 is 27 4

(a)

27 8

(b)

(c)

29 4

29 8

(d)

30. The value of Ú Ú 3 y dx dy over the triangle with vertices (–1, 1), (0, 0) and (1, 1) is [Winter 2015] (a) 0 (b) 1 (c) 2 (d) 3 31. The area of the curve y = x2 + 1 bounded by the x-axis and the line x = 1 and x = 2 is [Summer 2014] 3 10

(a)

10 3

(b)

(c) 6

1 6

(d)

32. The equation of a cylindrical surface x2 + y2 = 9 becomes ____ when converted to cylindrical polar coordinates. [Summer 2016] (a) r = 9 (b) r2 = 9 (c) r = ± 3 (d) r = 3 33.

2

x2

Ú0 Ú0

y x e

[Summer 2016]

dydx is equal to

(a) e2 – 1 (b) e2 2 2 1 34. Ú Ú dxdy = 1 1 xy (a) 0 (b) (log 2)2 35. The region of

4 6

Ú1 Ú2 dx dy

(a) rectangle (b) square 36. The region

2 2

Ú1 Ú1 dx dy

37. The value of (a) 0

Ú0 Ú0 (3 x (b) 1

2

(d) e–2 [Winter 2016]

(c) 1

(d) log 2 [Winter 2016]

represents (c) circle

(d) triangle [Summer 2017]

represents

(a) rectangle (b) square 1 1

(c) e2 + 1

(c) circle

(d) triangle

- 2 y 2 ) dx dy is (c) –1

[Summer 2017] (d)

1 3

Answers 1.(d) 2.(b) 3.(b) 4.(a) 5.(b) 6.(d) 7.(a) 8.(c) 11.(d) 12.(a) 13.(a) 14.(a) 15.(a) 16.(c) 17.(a) 18.(a) 21.(b) 22.(c) 23.(a) 24.(b) 25.(b) 26.(a) 27.(d) 28.(c) 31.(b) 32.(d) 33.(a) 34.(b) 35.(a) 36.(a) 37.(d)

9.(b) 10.(d) 19.(a) 20.(a) 29.(a) 30.(c)

1 Differential Appendix

Formulae 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

d n (x ) = nx n–1 dx d 1 (log x) = dx x d x (e ) = e x dx d x (a ) = ax log a dx d (sin x) = cos x dx d (cos x) = –sin x dx d (tan x) = sec2 x dx d (cot x) = –cosec2 x dx d (sec x) = sec x tan x dx d (cosec x) = –cosec x cot x dx d 1 (sin–1 x) = dx 1 − x2 d 1 (cos–1 x) = dx 1 - x2 d (tan–1 x) = 1 + 12 dx x

d 1 (cot–1 x) = − dx 1 + x2 1 d 15. (sec–1 x) = dx x x2 − 1 d 1 16. (cosec–1 x) = − dx x x2 − 1 14.

d dv du (uv) = u +v dx dx dx du dv −u v d  u d x dx 18.   = 2 dx  v  v 17.

Appendix

Integral Formulae 1.

∫ x dx n

=

xn +1 (n π –1) n +1

1

2.

∫ x dx = log x

3.

∫ e dx = e

4.

x ∫ a dx =

5.

∫ sin x dx = – cos x

6.

∫ cos x dx = sin x

7.

∫ tan x dx = –log cos x

8.

∫ cot x dx = log sin x

9.

∫ sec x dx = log (sec x + tan x) ∫ cosec x dx = log (cosec x – cot x) ∫ sec x dx = tan x

10. 11.

x

x

ax , a > 0, a π 1 log a

2

13.

∫ cosec x dx = – cot x ∫ sec x tan x dx = sec x

14.

∫ cosec x cot x dx = –cosec x

15.

x d x = sin–1   2 2  a a −x

12.

2

∫

1

2 16.

1

∫

x − a2 2

dx =

 x log ( x + x 2 − a 2 ) = cosh–1   a 1 dx = 17. ∫ 2 x + a2  x log ( x + x 2 + a 2 ) = sinh–1   a 1 1  a + x dx = log  − x2 2a  a − x  1 −1  x  = tanh   , x2 < a2 a a 1 1  x − a dx = 19. ∫ 2 log  x − a2 2a  x + a  18.

∫a

2

= −

1  x coth −1   , x2 > a2  a a

20.

∫a

2

1 1  x d x = tan −1    a a + x2

21.

∫

a 2 − x 2 dx

= 22.

x 2 a2  x sin −1   a − x2 +  a 2 2

∫

a 2 + x 2 dx

=

x 2 2 a2 a +x + log ( x + x 2 + a 2 ) 2 2

Integral Formulae

A2.2

23.

∫ =

24.

25.

26. 27.

29. 30.

x 2 2 a2 x −a − log ( x + 2 2

∫e

x2 − a2 )

∫e

31.

∫ sin[ f ( x)] f ′( x) d x = – cos f (x)

32.

∫ cos[ f ( x)] f ′( x) d x = sin f (x) a

33.

ax

sin bx d x e ax = 2 (a sin bx – b cos bx) a + b2

∫ f (a − x) d x 0

2a

34.

∫

f ( x) d x

0

ax

 du



∫ uv dx = u ∫ v d x − ∫  dx ∫ v d x dx

∫ [ f ( x)]

n

∫

∫

a

f ( x) d x =

0

cos bx d x e ax = 2 (a cos bx + b sin bx) a + b2

= 28.

x 2 − a 2 dx

a

=

0

0

a

35.

∫

f ( x) d x

−a a

= 2 ∫ f ( x) d x ,

f ′ ( x) d x

if f (x) is even

0

[ f ( x)]n +1 , n π –1 n +1 f ′ ( x) d x = log | f (x) | f ( x)

= 0, 36.

∫

f ( x) d x

0

a

= 2 Ú f ( x) d x,

f (x)

x

if f (x) is odd

2a

∫ e f ′ ( x) d x = e ∫ e [ f ( x) + f ′( x)] dx = e f (x) f ( x)

∫

a

f ( x ) d x + ∫ f ( 2a − x ) d x

if f (x) = f (2a – x)

0

= 0,

x

if f (x) = –f (2a – x)

reduction formulAe 1.

∫ sin

n

1 cos x sinn –1 x n n −1 n−2 + ∫ sin x dx

x dx = –

5.

tan x sec n − 2 x n −1 n−2 n−2 + ∫ sec x dx

n ∫ sec x dx =

n

1 2. ∫ cos n x dx = sin x cosn –1 x n n −1 n−2 + ∫ cos x dx n n −1 tan x 3. tan n x dx = ∫ n −1 – ∫ tan n − 2 x dx n −1

4.

∫ cot

n

cot x n −1 – ∫ cot n − 2 x dx

x dx = −

n −1

6.

∫ cosec x dx n

=

− cot x cosec n − 2 x n −1 n−2 + cosec n − 2 x dx n −1 ∫

7. (a)

∫ sin

m

x cos n x dx cos n + 1 x sin m −1 x = − m+ n m −1 + sin m − 2 x cos n x dx m+n ∫

Integral Formulae

(b)

(c)

(d)

∫ sin

x cos n x dx sin x cos n −1 x = m+ n n −1 + sin m x cos n − 2 x dx m+ n ∫

∫ sin

m

m +1

x cos n x dx cos n + 1 x sin m −1 x =− n +1 m −1 + sin m − 2 x cos n + 2 x dx n +1 ∫

∫ sin

=

n −1 n − 3 n − 5 1 p ◊ ◊ ◊…◊ ◊ , n−2 n−4 n 2 2 if n is even

9.

m

x cos n x dx sin m + 1 x cos n −1 x = m +1 n −1 + sin m + 2 x cos n − 2 x dx m +1 ∫

∫

cos n x dx n−5 n −1 n − 3 2 = ◊ ◊ ◊…◊ , n−2 n−4 n 3 0

=

n −1 n − 3 n − 5 1 p ◊ ◊ ◊…◊ ◊ , n−2 n−4 n 2 2 if n is even

10.

∫ =

∫

m+ n+ 2 m +1

∫ sin

(f)

m

m+ 2 n ∫ sin x cos x dx

8.

∫

n

sin x dx

0

=

n −1 n − 3 n − 5 2 ◊ ◊ ◊ …◊ , n−2 n−4 n 3 if n is odd,

m−3 m −1 ◊ ◊…◊ m+n m+ n−2

if m is odd and n may be odd or even =

m −1 m−3 m−5 ◊ ◊ m+ n m+ n−2 m+ n−4 …

=

p 2

sin m x cos n x dx

1 2 ◊ , 3 + n n +1

x cos n x dx

sin m + 1 x cos n + 1 x m +1 m+ n+ 2 + sin m x cos n + 2 x dx m +1 ∫

p 2

0

(e) sin m x cos n x dx

+

p 2

if n is odd

m

cos n + 1 x sin m + 1 x =− m +1

A2.3

1 n −1 n − 3 2 × ◊ … , 2+n n−2 n 3

if m is even and n is odd =

m −1 m − 3 m−5 ◊ … m+ n m+ n−2 m+ n−4 1 n −1 n − 3 1 p × ◊ … ◊ , 2+n n 2 2 n−2

if m is even and n is even

Appendix

Standard Curves Rectangular hyperbola xy = a

3 Catenary y = c cosh

( cx )

y

y x

O

c x

O

Cissoid of Diocles y2 (2a – x) = x3 y

Cubical parabola a2y = x3

x = 2a Asymptote

y x

O

(2a, 0)

O

Semi-cubical parabola ay2 = x3

x

ay2 = x(a2 + x2) y

y

O

x

O

x

A3.2

Standard Curves

Witch of Agnessi xy2 = 4a2 (a – x)

x (x2 + y2) = a(x2 – y2) y

y=

x

y

x O

A (a, 0)

x

O

y=

y(x2 + 4a2) = 8a3

�x

2y2 = x(4 + x2) y

y (0, 2a)

O

x

(4, 0) (2, 0)

x

O

a2 y2 = x3 (2a – x)

y2 = x5 (2a – x) y

y

(2a, 0) O

x

y2 (a2 – x2) = a3x

O

(2a, 0)

x

Cycloid x = a (q + sin q ), y = a (1 – cos q)

y y

x O

2a O q = –p

q=p q=0

x

Standard Curves

A3.3

Cycloid x = a (q + sin q ), y = a (1 + cos q ) Cycloid x = a (q – sin q ), y = a (1 + cos q ) y y

�p

x

p

O

q=�p

x

O q=0

q=p 2

q = 2p

q=p

2

2

Cycloid x = a (q – sin q), y = a (1 – cos q) Astroid x 3 + y 3 = a 3 y

y

(0, a)

x

(a, 0)

O

x

q = 2p

q=0 q=p 2

2

Ê xˆ 3 Ê yˆ 3 Hypocycloid Á ˜ + Á ˜ = 1 Ë a¯ Ë b¯

Folium of Descartes x3 + y3 = 3axy y

y (0, b)

O

x + y = 3a

x y= (3a/2, 3a/2)

3a

x

x

O

(0, a) x+y+a=0

Cardioid r = a(1 + cos q ) q=

O

Cardioid r = a(1 – cos q )

p 2

q=

q=0

O

p 2

q=0

A3.4

Standard Curves

Cardioid r = a(1 + sin q ) q=

Cardioid r = a(1 – sin q )

p 2

q=

p 2

O

q=0

O

Lemniscate of Bernoulli r2 = a2 cos 2q

r2 = a2 sin 2q q=

q=

q=0

p 2

p 2

O

q=0

q=0 O

Hyperbolic spiral rq = a

r = aq mq

q=

p 2

q=0 O

Spiral of Archimidies r = aq q=

O r = aq

p 2

r 2q = a2 q=

q=0

q=0

p 2 q=0

Standard Curves

Four leaved rose r = a cos 2q , a > 0

A3.5

Three leaved rose r = a sin 3q , a > 0 q=p 2

q=p 2

q=p 6

q = 5p 6 q=0

O

q=0

GUJARAT TECHNOLOGICAL UNIVERSITY BE SEMESTER-1st/2nd (New) Examination – Winter 2016 Subject Code: 2110014

Date: 24/01/2017

Subject Name: Calculus Q.1 Multiple-Choice Questions (a) Choose the appropriate answer out of the four options given in each of the following questions. [07] Ê p 1ˆ 1. The sequence sin Á + ˜ converges to Ë 6 n¯ (a) 0 (b) 1 (c) –1

(d) 0.5

Ê p 1ˆ Solution: The sequence sin Á + ˜ converges to 0.5. Ë 6 n¯ Ans. (d) •

2. The sum of the series (a)

p p -e

Ê eˆ  ÁË p ˜¯ n =1

(b)

n

is

e p -e

a = Solution: Sum of the series = 1- r Ans. (b) 3. The value of lim

x Æ6

(a) 0

sin( x - 6) is x-6 (b) 1

(c)

p e-p

e p e 1p

=

(d) e p

e pp

=

(c) –1

˘ 1 È ( x - 6) ( x - 6)5 + + ˙ ( x - 6) Í x Æ6 x - 6 Î 3! 5! ˚ lim

2 4 È ˘ = lim Í1 - ( x - 6) + ( x - 6) + ˙ x Æ6 Î 3! 5! ˚ =1–0+0 =1

Ans. (b)

e p -e

(d) 0.5 3

Solution

e p

Solved Question Papers Winter 2016

SQP.2

4. Asymptote parallel to y-axis of the curve y =

x2 is the line x-3

(a) x = 0 (b) y = 3 (c) x = 3 (d) does not exist Solution: Asymptotes parallel to y-axis are obtained by equating the coefficient of highest degree term of y in the equation to zero. Equating the coefficient of highest degree term of y i.e. x – 3 to 0, x–3=0 x=3 Ans. (c) 5. Let f(x, y) = y sin(xy). The value of fx(p, 1) is (a) 0 Solution:

(b) 1

(c) –1

(d) 2.5

f(x, y) = y sin(xy) fx(x, y) = y2 cos(xy) fx(p, 1) = (1)2 cos(p) = –1

Ans. (c) 22

1

Ú Ú xy dxdy =

6.

11

(a) 0

(b) (log2)2

(c) 1

(d) log2

Solution: 2 2 1È 1 ˘ 2 1 d y = d x Í ˙ Ú y ÍÚ x ˙ Ú log x 1 ◊ y dy 1 1 Î1 ˚ 2

= log x

2 1

log y

2 1

= (log2 – log1)2 = (log2)2 Ans.

(b)

7. The coefficient of x5 in the expansion of ex is (a)

1 5

(b)

1 4!

(c)

Solution: Expansion of ex = 1 + x + The coefficient of x5 in ex = Ans. (c)

1 5!

1 5!

(d) 5

x2 x3 x 4 x5 + + + + 2! 3! 4! 5!

Solved Question Papers Winter 2016

SQP.3

(II) Choose the appropriate answer out of the four options given in each of the following questions. •

Â

1.

n =1

2n is 3n - 1

(a) convergent and sum is 0 (c) divergent Solution:

un =

2n 3n - 1

un +1 =

2 n +1 3n + 2

(b) convergent and sum is 1 (d) oscillating

un 2n 3n + 2 = ◊ un +1 3n - 1 2 n +1 Ê n 3+ un 1 ÁË lim = lim ◊ n Æ• un + 1 n Æ• 2 Ê nÁ3 Ë

2ˆ ˜ n¯ 1ˆ ˜ n¯

1 0, fyy = t = 2, fxy = s = 0 rt – s2 = 4 – 0 = 4 > 0 2 rt – s > 0 and r > 0. Hence, f(x, y) is minimum at point (0,0). Minimum value f(x, y) = f(0, 0) = 0 + 0 = 0 Ans. (d) 7. If x = u + 3v and y = –u + v then J = (a) –1

Solution:

Ans. (b)

(b) 4 ∂x ∂( x, y) ∂u J= = ∂(u, v) ∂y ∂u

∂( x, y) is ∂(u, v) (c) 5

∂x ∂v = 1 3 = 1 + 3 = 4 -1 1 ∂y ∂v

(d) 7

p 2

Solved Question Papers Winter 2016

SQP.5

Q2 (a) You drop a ball from ‘a’ meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh where 0 < r < 1. Find 2 the total distance the ball travels up and down, when a = 6 m and r = m. 3 Solution: Refer Example 6 on page 1.14. Q2 (b) Evaluate i. lim

xÆ0

e x + e- x - x 2 - 2 sin 2 x - x 2

Solution: Refer Example 5 on page 4.5. ii. lim Ê x - 1 ˆ Á x Æ1 Ë x - 1 log x ˜¯ Solution: Refer Example 3 on page 4.39. Q2 (c) Obtain the Maclaurin’s series of loge(1 + x) and hence find the series of Ê1 + xˆ Ê 11ˆ . Also obtain the approximate value of loge Á ˜ . loge Á ˜ Ë 9¯ Ë1- x¯ Solution: Refer Example 2 on page 2.31. Ï 2 x2 y Ô Q3 (a) Show that f ( x, y) = Ì x 3 + y3 Ô 0 Ó is not continuous at the origin.

( x, y ) π 0 ( x, y ) = 0

Solution: Refer Example 11 on page 7.8. Q3 (b) If q = t

n

-r 2 e 4t

then find the value of n for which

1 ∂ Ê 2 ∂q ˆ ∂q . Ár ˜= r 2 ∂r Ë ∂r ¯ ∂t

Solution: Refer Example 32 on page 7.39. Q3 (c) State Euler’s theorem on homogenous function of two variables. If Ê x 2 + y2 ˆ u = tan -1 Á , prove that Ë x + y ˜¯ x2

∂2 u ∂x 2

+ 2 xy

∂2 u ∂2 u + y 2 2 = - 2 sin 3 u cos u ∂x ∂y ∂y

Solution: Refer Example 22 on page 7.153.

Solved Question Papers Winter 2016

SQP.6

Q4 (a) Find the Jacobian the transformation x = r sin f cos q,

y = r sin f sin q,

z = r cos f

Solution: Refer Example 4 page 8.103. Q4 (b) Find the equations of the tangent plane and normal line to the surface f(x, y, z) = x2 + y2 + z – 9 = 0 at the point p(1, 2, 4). Solution: Refer Example 2 on page 8.2. Q4 (c) A rectangular box open at the top is to have a volume of 32 cubic units. Find the dimensions of the box requiring least material for its construction. Solution: Refer Example 11 on page 8.60. 1 2

Q5 (a) Evaluate

Ú Ú (1 - 6 x

2

y)dxdy.

-1 0

Solution: Refer Example 3 on page 9.4. 1 z 2p

Q5 (b) Evaluate

Ú Ú Ú (r

2

cos2 q + z 2 )rdq drdz .

0 0 0

Solution: Refer Example 4 on page 9.114. Q5 (c) Change the order of integration and evaluate 2 1 1- x

Ú Ú 0

ey (e y + 1) 1 - x 2 - y 2

0

dy dx

Solution: Refer Example 8 on page 9.49. Q6 (a) Let S =

•

 na n where |a | < 1. Find the value of a in (0, 1) such that S = 2a.

n =1

Solution: Refer Example 4 on page 1.13. Q6 (b) Check for convergence/divergence i.

•

5n3 - 3n

n =1

n2 (n - 2)(n2 + 5)

Â

.

Solution: Refer Example 19 on page 1.29. •

ii.

2n

 n3 + 1. n =1

Solution: Refer Example 3 on page 1.39.

Solved Question Papers Winter 2016

•

Q6 (c) 1. Test the series for absolute or conditional convergence Solution: Refer Example 3 on page 1.86.

Â

n =1

SQP.7

(-1)n . n + n +1

OR •

Q6 (c) 2. For the series

 (-1)n -1

n =1

x 2 n -1 , find the radius and internal of convergence. 2 n -1

Solution: Refer Example 6 on page 1.100. Q7 (a) The graph of y = x2 between x = 1 and x = 2 is rotated around the x-axis. Find the volume of the solid so generated. Solution: Refer Example 3 on page 6.9. Q7 (b) Test the convergence of the improper integrals. If convergent, then evaluate the same. 1

i.

dx

Ú1- x 0

Solution: Refer Example 3 on page 5.11. •

ii.

1

Ú (1 + x 2 )(1 + tan -1 x) dx 0

Solution: Refer Example 10 on page 5.6. Q7 (c) Trace the curve y2(a + x) = x2(a – x) a > 0. Solution: Refer Example 4 on page 3.8.

GUJARAT TECHNOLOGICAL UNIVERSITY BE Semester-1st/2nd (New) Examination – Summer 2017 Subject Code: 2110014 Subject Name: Calculus

Date: 01/06/2017

Q.1 Multiple-Choice Questions (I) Choose the appropriate answer out of the four options given in each of the following questions. 1. Infinite series 1 +

1 1 1 + 2 + 3 + ... is 2 2 2

(a) divergent

(b) convergent

Solution:

1+

(c) oscillatory

(d) none of these

1 1 1 + + + 2 2 2 23

a = 1, r =

1 , |r |0 f ( x) = Ì Ó- x x < 0 lim f ( x ) = lim x = 0

x Æ 0+

x Æ 0+

lim f ( x ) = lim (- x ) = 0

x Æ 0-

x Æ 0-

Ans. (a) 6. Curve y2(a + x) = x2(b – x) is symmetric about _____ (a) x-axis (b) y-axis (c) line x = b (d) line x = –a Solution: The powers of y occurring in the equation are all even. Hence, the curve is symmetric about x-axis. Ans. (a) 7. The curve increases strictly in interval in which (a) < 0 Solution:

(b) > 0

dy is dx

(c) = 0

(d) none of these

dy >0 dx

Ans. (b) (b) Choose the appropriate answer out of the four options given in each of the following questions. •

1. The value of

Úe

-x

cos(2 x )dx is

0

(b) -

(a) 0

1 5

(c)

1 5

(d)

•

Solution:

Èb -x ˘ -x = e cos2 x d x lim Í Ú e cos(2 x )dx ˙ Ú b Æ• Í ˙˚ 0 Î0 b

ÔÏ e x Ô¸ (- cos2 x + 2sin 2 x ) ˝ = lim Ì b Æ• Ô 1 + 4 Ô˛0 Ó

2 5

Solved Question Papers Summer 2017

SQP.3

ÏÔ e - x ¸Ô 1 = lim Ì (- cos 2 x + 2 sin 2 x ) ˝ - (-1) b Æ• Ô 5 Ó ˛Ô 5 =

1 5

Ans. (a) 22

2. What does the region

Ú Ú dxdy

represent?

11

(a) Rectangle (b) Square (c) Circle (d) Triangle Solution: All four limits are constant. Hence, the region represents a rectangle. Ans. (a) Ê cos x ˆ 3. The value of lim Á p ˜ is p x˜ xÆ Á 2¯ 2 Ë (a) 0 Solution:

(b) 1 lim

p xÆ 2

(c) –1

(d)

p 2

Ê0 ˆ ÁË form˜¯ 0

cos x p x2

Ê sin x ˆ = lim Á ˜ pË 1 ¯ xÆ

(L’Hospital’s rule)

2

= lim (- sin x ) xÆ

p 2

Êpˆ = - sin Á ˜ Ë 2¯ = -1 Ans. (c) Ê yˆ 4. The function f(x, y) = x2y f Á ˜ is homogenous of degree Ë x¯ (a) 0 (b) 1 (c) 2 (d) 3 Ê yˆ f ( x, y = x 2 y f Á ˜ Ë x¯ Replacing x by xt and y by yt, Solution:

Ê yˆ f ( xt , yt ) = t 3 x 2 y f Á ˜ Ë x¯

SQP.4

Solved Question Papers Summer 2017

= t3 f(x, y) which is a homogenous function of degree 3. Ans. (d) 5. The equation of the form f(xy) = c then (a) -

fx fy

Solution:

(b)

fx fy

dy = _______ dx (c) -

fy

(d)

fx

fy fx

∂f f dy = - ∂x = - x ∂f dx fy ∂y

Ans. (a) 11

6. The value of

Ú Ú (3 x

2

- 2 y 2 )dx dy is

00

(a) 0

(b) 1

(c) –1

(d)

1 3

1 È1 ˘ 2 2 2 2 (3 x 2 y )d x d y = ÚÚ Ú ÍÍÚ (3 x - 2 y )dx ˙˙ dy 00 0 Î0 ˚ 11

Solution:

1

3 = Ú 3 x - 2 xy 2 3 0

1

dy 0

1

= Ú (1 - 2 y 2 )dy 0

3 = y - 2y 3

1

0

2 =13 1 = 3 Ans. (d) 7. If x = u + 3v and y = v – u then the value of (a) 4

(b) –1

∂( x, y) ∂(u, v)

(c) 5

(d) 7

Solved Question Papers Summer 2017

Solution:

∂x ∂( x, y) ∂u = ∂y ∂(u, v) ∂u

SQP.5

∂x ∂v = 1 3 = 1 + 3 = 4 -1 1 ∂y ∂v

Ans. (a) Q2 (a) Expand log(sec x) in power of x. Solution: Refer Example 1 on page 2.57. 1

Ê 1x + 2 x + 3 x ˆ x Q2 (b) Evaluate lim Á ˜¯ xÆ0 Ë 3

Solution: Refer Example 5 on page 4.48. Q2 (c) (i) Trace the curve y2(2a – x) = x3. Solution: Refer Example 2 on page 3.6. 3

(ii) Determine

Ú 0

1 dx converge or diverges. 3- x

Solution: Refer Example 1 on page 5.10. Q3 (a) If f(x, y) = x2y + xy2 then find fx(1, 2) and fy(1, 2) by definition. Solution: Refer Example 1 on page 7.12. Q3 (b) Check the continuity for the following function at (0, 0). Ï 2 xy Ô f ( x, y ) = Ì x 2 + y 2 Ô 0 Ó

( x, y) π (0, 0) ( x, y) = (0, 0)

Solution: Refer Example 10 on page 7.7. 1 ˆ Ê 1 4 + y4 x Á ˜ Q3 (c) (i) If u = sin -1 Á 1 1 ˜ ÁË x 5 + y 5 ˜¯

x2

then find the values of x

2 ∂2 u ∂2 u 2 ∂ u + 2 xy + y . ∂x ∂y ∂x 2 ∂y 2

Solution: Refer Example 9 on page 7.142.

∂u ∂u +y ∂x ∂y

and

SQP.6

Solved Question Papers Summer 2017

Q3 (c) (ii) If u =

ex + y + z ∂u ∂u ∂u then show that + + = 24 . e x + e y + ez ∂x ∂y ∂z

Solution: Refer Example 4 on page 7.14. Q4 (a) Find the extreme value of x3 + 3xy2 – 3x2 – 3y2 + (4), if any. Solution: Refer Example 5 on page 8.25. Q4 (b) Find the equation of the tangent plane and normal line to the surface 2x2 + y2 + 2z = 3 at (2, 1, –3). Solution: Refer Example 4 on page 8.3. Q4 (c) (i) Find a point on the plane 2x + 3y – z = 5 which is nearest to the origin. Solution: Refer Example 4 on page 8.41. Q4 (c) (ii) Expand exy in the power of x – 1 and y – 1 using Taylor’s expansion. Solution: Refer Example 4 on page 8.84. Q5 (a) Test the convergence of the series

1 x x2 + + + . 1◊ 2 ◊ 3 4 ◊ 5 ◊ 6 7 ◊8 ◊ 9

Solution: Refer Example 25 on page 1.56. •

Q5 (b) Test the convergence of the series

1

 [(n3 + 1) 3 - n] .

n =1

Solution: Refer Example 22 on page 1.31. Q5 (c) (i) Determine absolute or conditional convergence of the series •

 (-1)n

n =1

n2 n3 + 1

Solution: Refer Example 4 on page 1.87. (ii) Test the convergence for 5 -

10 20 40 + + . 3 9 47

Solution: Refer Example 2 on page 1.12. Q6 (a) Evaluate

ÚÚ r

a 2 - r 2 drdq over the upper half of the circle r = a cos q.

Solution: Refer Example 2 on page 9.72.

Solved Question Papers Summer 2017

4

Q6 (b) Sketch the region of integration and evaluate

x

ÚÚ 1 0

3 e 2

SQP.7

y x

dy dx.

Solution: Refer Example 12 on page 9.9. 21

Q6 (c) (i) Evaluate the integral

ÚÚe

x2

dxdy by changing the order of integration.

0 y 2

Solution: Refer Example 5 on page 9.46. 2 2 yz

(ii) Evaluate the integral

Ú Ú Ú xyz dxdydz. 01 0

Solution: Refer Example 10 on page 9.118. Q7 (a) Find the area included between the curve y2(2a – x) = x3 and its asymptote. Solution: Refer Example 10 on page 9.149. Q7 (b) Find the volume of the solid generated by revolving the cardioid r = a(1 – cos q) about the initial line. Solution: Refer Example 1 on page 6.25. Q7 (c) Use triple integration to find the volume of the solid within the cylinder x2 + y2 = 9 between the planes z = 1 and x + z = 1. Solution: Refer Example 9 on page 9.180.

Index

A Absolute Convergence of a Series 1.84 Alternating Series 1.77 Area as Double Integral 9.141 Area in Cartesian Coordinates 9.141 Area in Polar Coordinates 9.154

Convergence and Divergence of Improper Integrals 5.16 Convergence, Divergence and Oscillation of a Sequence 1.1 Convergence, Divergence and Oscillation of Infinite Series 1.8

D B Bounded Sequence 1.2

C Cauchy’s Integral Test 1.71 Cauchy’s Root Test 1.63 Chain Rule 7.70 Change of Order of Integration 9.31 Change of Variables 9.77 Change of Variables from Cartesian to Other Coordinates 9.95 Change of Variables from Cartesian to Polar Coordinates 9.77 Clairaut’s Theorem 7.12 Comparison Test 1.17 Composite Function of One Variable 7.70 Composite Function of Two Variables 7.77 Concavity, Convexity and Points of Inflection of a Curve 3.1 Conditional Convergence of a Series 1.84

D’Alembert’s Ratio Test 1.36 Deductions from Euler’s Theorem 7.115 Direct Comparison Test 5.17 Double Integrals in Polar Coordinates 9.66 Double Integrals Over Rectangles 9.1

E Errors and Approximations 8.12 Euler’s Theorem for a Function of Three Variables 7.114 Euler’s Theorem for a Function of Two Variables 7.114 Euler’s Theorem for Homogeneous Functions 7.114 Evaluation of Double Integrals by Fubini’s Theorem 9.2

F First Derivative Test for Local Extremum 3.2 Functions of Two or More Variables 7.1

I.2

Index

G Geometric Series 1.10 Graphs and Level Curves 7.1

H Higher-Order Partial Derivatives 7.11

I Implicit Differentiation 7.105 Improper Integral of the Third Kind 5.16 Improper Integrals 5.1 Improper Integrals of the First Kind 5.1 Improper Integrals of the Second Kind 5.9 Infinite Series 1.8 Interval and Radius of Convergence 1.94

J Jacobians 8.97

L L’Hospital’s Rule 4.1 Leibnitz’s Test for Alternating Series 1.77 Limit and Continuity of Functions of Several Variables 7.2 Limit Comparison Test 5.18 Limit of a Sequence 1.1 Linear Approximation or Linearization 8.10

M Maclaurin’s Series 2.27 Maxima and Minima 3.2

Maximum and Minimum Values by Second Derivative Test 8.21 Maximum and Minimum Values with Constrained Variables 8.38 Method of Lagrangian Multipliers 8.49 Monotonic Functions 3.1 Monotonic Sequence 1.2

N nth Term Test for Divergence 1.9

P Partial Derivatives 7.11 Power Series 1.94 Properties of Double Integrals 9.3 Properties of Infinite Series 1.8 Properties of Jacobians 8.98

R Relation between Polar and Cartesian Coordinates 3.31

S Sandwich Theorem for Sequences 1.6 Second Derivative Test for Local Extremum 3.3 Sequence 1.1

T Tangent Plane and Normal to a Surface 8.1 Taylor’s Formula for Two Variables 8.81 Taylor’s Series 2.1 Total Derivatives 7.70 Tracing of Cartesian Curves 3.4 Tracing of Parametric Curves 3.24

Index

Tracing of Polar Curves 3.31 Triple Integrals 9.109 Triple Integrals in Cartesian Coordinates 9.109 Triple Integrals in Cylindrical Coordinates 9.109 Triple Integrals in Spherical Coordinates 9.110

I.3

Uniform Convergence of a Series 1.85

Volume by Slicing 6.1 Volume of Solid of Revolution 6.6 Volume of Solid of Revolution in Cartesian Form 6.6 Volume of Solid of Revolution in Parametric Form 6.7 Volume of Solid of Revolution in Polar Form 6.7 Volume of Solids of Revolution by Disk Method 6.6 Volume of Solids of Revolution by Washer Method 6.6

V

W

Volume as Triple Integral 9.169 Volume by Cylindrical Shells 6.33

Weierstrass’s M-Test 1.91

U