Fondamenti e complementi di Analisi matematica 2 [1]

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Fondamenti e complementi di Analisi matematica 2 A cura di Carlo Mariconda Corsi di Laurea Triennale in Ingegneria Biomedica ed Elettronica Corso di Laurea Triennale in Ingegneria Informatica Università degli Studi di Padova

Joel Hass Maurice D. Weir George B. Thomas, Jr.

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FONDAMENTI E COMPLEMENTI DI ANALISI MATEMATICA 2

Fondamenti e complementi di Analisi matematica 2 a cura di Carlo Mariconda Corsi di Laurea Triennale in Ingegneria Biomedica ed Elettronica Corso di Laurea Triennale in Ingegneria Informatica Università degli Studi di Padova

Joel Hass Maurice D. Weir George B. Thomas, Jr.

© 2019 Pearson Italia, Milano – Torino

ESTRATTO DAL VOLUME: Joel Hass, Maurice D. Weir, George B. Thomas Jr., Analisi matematica 2

Authorized translation from the English language edition, entitled UNIVERSITY CALCULUS: EARLY TRASCENDENTALS, MULTIVARIABLE 2nd edition, by J. Hass, M.D. Weir, G.B. Thomas Jr., published by Pearson Education, Inc, publishing as Pearson, Copyright © 2012. All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage retrieval system, without permission from Pearson Education, Inc. Italian language edition published by Pearson Italia S.p.A., Copyright © 2014.

Le informazioni contenute in questo libro sono state verificate e documentate con la massima cura possibile. Nessuna responsabilità derivante dal loro utilizzo potrà venire imputata agli Autori, a Pearson Italia S.p.A. o a ogni persona e società coinvolta nella creazione, produzione e distribuzione di questo libro. Per i passi antologici, per le citazioni, per le riproduzioni grafiche, cartografiche e fotografiche appartenenti alla proprietà di terzi, inseriti in quest’opera, l’editore è a disposizione degli aventi diritto non potuti reperire nonché per eventuali non volute omissioni e/o errori di attribuzione nei riferimenti.

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Impaginazione: Andrea Astolfi Immagine di copertina: photonic 8 / Alamy Stock Photo Grafica di copertina: Maurizio Garofalo Stampa: Rotomail, Vignate (MI) Tutti i marchi citati nel testo sono di proprietà dei loro detentori.

9788891915085 Printed in Italy 1a edizione: settembre 2019 Ristampa

Anno

00 01 02 03 04

19 20 21 22 23

Sommario

Capitolo 1 Equazioni parametriche e coordinate polari 1.1 1.2 1.3 1.4 1.5

Parametrizzazione delle curve nel piano Analisi matematica con le curve parametriche Coordinate polari Tracciare grafici in coordinate polari Aree e lunghezze in coordinate polari

Capitolo 2 Funzioni a valori vettoriali e moto nello spazio 2.1 2.2 2.3 2.4

Curve nello spazio e loro tangenti Integrali di funzioni vettoriali e moto del proiettile Lunghezza d’arco nello spazio Curvatura e vettori normali a una curva

Capitolo 3 Derivate parziali 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Funzioni di più variabili Limiti e continuità in più dimensioni Derivate parziali La regola della catena Derivate direzionali e vettori gradiente Piani tangenti e differenziali Valori estremi e punti di sella

Capitolo 4 Integrali multipli 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Integrali doppi e iterati su rettangoli Integrali doppi su regioni arbitrarie Aree calcolate con una doppia integrazione Integrali doppi in coordinate polari Integrali tripli in coordinate rettangolari Momenti e centri di massa Integrali tripli in coordinate cilindriche e sferiche Sostituzione negli integrali multipli

1 1 6 12 15 18

31 31 39 43 48

65 65 72 78 90 95 103 110

135 135 140 147 149 154 162 168 177

]P  :VTTHYPV

Capitolo 5 Integrazione sui campi vettoriali 5.1 Integrali curvilinei 5.2 Campi vettoriali e integrali curvilinei: lavoro, circolazione e flusso 5.3 Indipendenza dai cammini, campi conservativi e potenziali 5.4 Il Teorema di Green nel piano 5.5 Superfici e aree 5.6 Integrali di superficie 5.7 Il Teorema di Stokes 5.8 Il Teorema della divergenza e una teoria unificata

Capitolo 6 Equazioni differenziali del primo ordine 6.1 Equazioni differenziali del primo ordine e problemi ai valori iniziali 6.2 Equazioni differenziali a variabili separabili 6.3 Equazioni lineari del primo ordine 6.4 Equazioni riconducibili a equazioni lineari o a variabili separabili 6.5 Applicazioni 6.7 Soluzioni grafiche di equazioni autonome

189 189 194 205 214 224 232 240 251

267 267 272 280 285 288 297

Complementi al testo Appendice – Formula di Taylor per funzioni di due variabili Risposte agli esercizi dispari

* ( 7 0 ; 6 3 6



Equazioni parametriche e coordinate polari PANORAMICA

Sommario del capitolo  Parametrizzazione delle curve nel piano  Analisi matematica con le curve parametriche  Coordinate polari  Tracciare grafici in coordinate polari  Aree e lunghezze in coordinate polari  Coniche in coordinate polari

In questo capitolo studieremo dei metodi per definire le curve nel piano. Anziché pensare a una curva come il grafico di una funzione, la considereremo più in generale come il percorso di una particella la cui posizione varia nel tempo. Le coordinate x e y della posizione della particella diventano così funzioni di una terza variabile t. Si può anche scegliere di descrivere i punti del piano in un nuovo modo, utilizzando le coordinate polari anziché il sistema di coordinate rettangolari o cartesiane. Questi nuovi strumenti sono utili nella descrizione del moto, come ad esempio nel caso di pianeti, satelliti o proiettili, che si muovono nel piano o nello spazio. Parabole, ellissi e iperboli descrivono le traiettorie di proiettili, pianeti e qualsiasi altro oggetto che si muova soggetto solamente alla forza gravitazionale o elettromagnetica.

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(a)

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0

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(c)

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(b) Rispetto all'asse y

Criteri di simmetria per grafici in coordinate polari

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Pendenza

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p

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2

1

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ESEMPIO 2

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⎛ 3 , 4⎛ ⎝2 3 ⎝

(c)

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(a) r 2 1

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r 2  sin 2

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r  1sin 2



0

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y

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U = VLQu 0

x

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Area nel piano y  S

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1 

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Funzioni a valori vettoriali e moto nello spazio PANORAMICA

Sommario del capitolo

In questo capitolo estenderemo lo studio delle curve piane, introdotte nel capitolo precedente, alle curve nello spazio, riguardandole come funzioni a valori vettoriali. Il dominio di tali funzioni è un intervallo di ℝ mentre il codominio è dato da un insieme di vettori, anziché di scalari. La parte dell’analisi matematica che tratteremo trova applicazione nella descrizione della traiettoria e del moto di oggetti nel piano o nello spazio, la cui velocità e la cui accelerazione lungo la traiettoria percorsa, come vedremo, sono rappresentate da vettori. Presenteremo inoltre la definizione di nuove quantità che descrivono la forma assunta dalla traiettoria di un oggetto nello spazio.

 Curve nello spazio e loro tangenti  Integrali di funzioni vettoriali e moto del proiettile  Lunghezza d’arco nello spazio  Curvatura e vettori normali a una curva  Componente tangenziale e componente normale dell’accelerazione  Velocità e accelerazione in coordinate polari

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DEFINIZIONE

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(4, 0, 0)

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Parametrizzazioni equivalenti &RPHDEELDPRYLVWRXQDFXUYDVLSXzSDUDPHWUL]]DUHLQGLYHUVLPRGLDVHFRQGD GHOODPRGDOLWjFRQFXLYLHQHSHUFRUVD1DWXUDOPHQWHOHSURSULHWjJHRPHWULFKHGHO ODFXUYDFKHqLOFRGRPLQLRGHOOHVXHSDUDPHWUL]]D]LRQLQRQSRVVRQRGLSHQGHUH GDOODSDUDPHWUL]]D]LRQHVFHOWD0DqDQFKHYHURFKHQRQWXWWHOHSDUDPHWUL]]D ]LRQLGLXQDGDWDFXUYDVRQRHTXLYDOHQWLQHOVHQVRFKHGHVFULYRQRODVWHVVDWUD LHWWRULDSHUFKpDGHVHPSLRXQDFXUYDFKLXVDSRWUHEEHHVVHUHSHUFRUVDSLYROWH ,QWURGXFLDPRDOORUDODVHJXHQWHGH¿QL]LRQH

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DEFINIZIONE

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UW = RXW  SHURJQLWH,



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Lunghezza d’arco lungo una curva nello spazio

Punto di riferimento 3

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Integrali multipli PANORAMICA

Sommario del capitolo

In questo capitolo consideriamo l’integrale di una funzione di due variabili ƒ(x, y) su una regione del piano e l’integrale di una funzione di tre variabili ƒ(x, y, z) su una regione dello spazio. Questi integrali multipli sono definiti come limiti di somme approssimanti di Riemann, proprio come gli integrali in una variabile. Mostreremo numerose applicazioni degli integrali multipli, tra cui il calcolo di volumi, aree, momenti e centri di massa.

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Integrazione sui campi vettoriali PANORAMICA

Sommario del capitolo  Integrali curvilinei  Campi vettoriali e integrali curvilinei: lavoro, circolazione e flusso  Indipendenza dai cammini, campi conservativi e potenziali  Il Teorema di Green nel piano  Superfici e aree

In questo capitolo estendiamo la teoria dell’integrazione alle curve e superfici nello spazio. La teoria degli integrali curvilinei e di superficie che ne risulta fornisce utilissimi strumenti per le scienze e l’ingegneria. Gli integrali curvilinei si usano per esempio per calcolare il lavoro compiuto da una forza mentre un oggetto si sposta lungo una traiettoria e per trovare la massa di un cavo curvo con densità non omogenea. Gli integrali di superficie si usano per trovare il flusso di un fluido attraverso una superficie. Presentiamo i teoremi fondamentali del calcolo integrale vettoriale e ne studiamo le conseguenze matematiche e le applicazioni fisiche. Per concludere, mostreremo che i teoremi principali sono interpretazioni generalizzate del teorema fondamentale del calcolo integrale.

 Integrali di superficie  Il Teorema di Stokes  Il Teorema della divergenza e una teoria unificata

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1

Note sul Capitolo 1

Definizione di curva parametrica nel piano (Sez. 1.1) Precisiamo meglio la definizione di curva parametrica. Definizione 1.1 (Curva parametrica) Una curva parametrica in R2 e` una coppia di funzioni continue

(f, g) : I → R2 ,

t → (f (t), g(t))

definite su un intervallo non degenere I di R. Diremo che t → (f (t), g(t)) e` una funzione vettoriale continua a valori in R2 . L’insieme dei punti (x, y) = (f (t), g(t)), al variare di t ∈ I , si chiama sostegno della curva parametrica, di cui (f, g) e` una parametrizzazione. Osservazione 1.2 Una curva parametrica nel piano non e` un insieme ma una coppia di funzioni continue definite su un intervallo. Per capire la differenza tra curva parametrica e il suo sostegno immaginiamo di fotografare al buio un punto luminoso, con l’otturatore aperto. La funzione che al tempo t associa la posizione del punto e` una curva parametrica. La scia luminosa ottenuta in fotografia rappresenta il sostegno della curva parametrica, la traiettoria istante per istante del punto luminoso ci fornisce la parametrizzazione (le componenti in t sono la posizione del punto all’istante t).



In tutto il Libro per curva si intende una curva parametrica: nell’uso comune capita spesso di confondere una curva parametrica con il suo sostegno, a sua volta chiamato spesso semplicemente “curva”. Il problema e` che un dato insieme pu`o essere il sostegni di infinite curve parametriche distinte. Quale curva parametrica si considera? Una parametrizzazione “naturale” o ad essa equivalente (si veda la definizione a pag. 44). Vediamo alcuni esempi al proposito. Esempio 1.3 (Il segmento) Il segmento da un punto P a un punto Q in Rn e` la curva parametrica γ(t) = P + t(Q − P ), t ∈ [0, 1]

4

Note sul Capitolo 1

o una sua parametrizzazione equivalente. Il suo sostegno e` il segmento P Q. Si noti che anche il segmento da Q a P dato da

γ −1 (t) = Q + t(P − Q), t ∈ [0, 1] ha lo stesso sostegno: in tal caso tuttavia esso viene percorso partendo da Q in t = 0 e arrivando a P in t = 1. Quando si parla del segmento P Q si sottointende nel corso di considerare la curva parametrica γ . Esempio 1.4 (Circolo unitario) Per circolo unitario si intende la curva parametrica γ1 (t) = (cos t, sin t), t ∈ [0, 2π], o una parametrizzazione equivalente, come ad esempio (cos(2t), sin(2t)), t ∈ [0, π]. Il sostegno del circolo unitario e` in cerchio unitario {(x, y) ∈ R2 : x2 + y 2 = 1}. Tale insieme e` anche il sostegno della curva che lo percorre 2 volte data da γ2 (t) = (cos t, sin t), t ∈ [0, 4π]. Si noti tuttavia che le curve parametriche γ1 , γ2 non sono equivalenti: infatti se lo fossero esisterebbe φ : [0, 2π] → [0, 4π] continua e crescente tale che γ1 (t) = γ2 (φ(t)) per ogni t ∈ [0, 2π], da cui

cos t = cos(φ(t)),

sin t = sin(φ(t)) ∀t ∈ [0, 2π],

da cui

φ(t) = t + 2k(t)π ∀t ∈ [0, 2π], con k(t) ∈ Z. Necessariamente k(t) e` continua, quindi costante: k(t) ≡ k ∈ Z. Da φ(0) = 2kπ si deduce che k = 0, ma allora φ(2π) = 2π = 4π , contraddicendo il fatto che φ([0, 2π]) = [0, 4π]. Un altro modo per provare l’affermazione consiste nel notare che la lunghezza di γ1 vale 2π , quella di γ2 vale 4π , mentre curve equivalenti hanno la stessa lunghezza. Quando si parla di “cerchio” si intende sempre una curva parametrica con parametrizzazione equivalente a γ1 . Vediamo alcuni esempi fondamentali di curve parametriche. Esempio 1.5 (Il segmento) Il segmento che congiunge due punti nel piano p = (p1 , p2 ) e q = (q1 , q2 ) e` una curva: infatti una sua parametrizzazione e` data da

r(t) = p + t(q − p) : t ∈ [a, b]. In tutto il corso con “segmento [p, q]” intenderemo la curva parametrica r (e non l’insieme di punti del segmento), o una ad essa equivalente. Esempio 1.6 (Il cerchio) Un cerchio di raggio R > 0 e centro p = (p1 , p2 ) e` una curva; una sua parametrizzazione e`

r(t) = (p1 + R cos t, p2 + R sin t),

t ∈ [0, 2π].

Note sul Capitolo 1

5

In tutto il corso con “cerchio di centro p e raggio R” intenderemo non la curva (cio`e l’insieme dei punti di tale cerchio) ma la parametrizzazione data sopra o una qualunque ad essa equivalente. Si noti che il cerchio percorso una volta al doppio della velocit`a, dato da

γ(t) = (p1 + R cos(2t), p2 + R sin(2t)),

t ∈ [0, π]

e` equivalente al cerchio di centro p e raggio R: infatti γ(t) = r(2t) per ogni t ∈ [0, π] e la funzione t → u(t) = 2t e` biiettiva da [0, π] su [0, 2π], derivabile, con derivata continua e diversa da 0. Si pu`o dimostrare che il cerchio di stesso centro e raggio percorso pi`u volte, come ad esempio

(p1 + R cos t, p2 + R sin t),

t ∈ [0, 4π],

non e` equivalente a r. Esempio 1.7 (Il grafico di una funzione) Il grafico C = {(t, h(t)) : t ∈ [a, b]} di una funzione continua h : [a, b] → R e` una curva. Infatti la funzione vettoriale











 Figura 1.1. Il grafico della funzione sin t su [0, 2π]

r : t ∈ [a, b] → (t, h(t)) ne e` una parametrizzazione. Diremo, sfruttando l’abuso di linguaggio descritto sopra, che tale curva parametrica e` la “curva” y = h(x), al variare di x ∈ [a, b]. Si noti che un’altra parametrizzazione di C e` data dalla funzione t ∈ [a−1, b−1] → (t+1, h(t+1)): vi sono infinite parametrizzazioni di C , ognuna di queste e` una diversa curva parametrica con stesso sostegno C .

Somme di Riemann (Sez. 1.2) Le somme di Riemann sono spesso utili in questo corso. Ne richiamiamo qui definizione e propriet`a. Sia g : [a, b] → R una funzione continua e sia a = t0 < t1 < ... < tm+1 = b una suddivisione di [a, b]. Una somma di Riemann di g associata alla suddivisione data e` una somma del tipo

g(t0 )(t1 − t0 ) + ... + g(tm )(tm+1 − tm ),

6

Note sul Capitolo 1

con ti ∈ [ti , ti+1 ] per ogni 0 ≤ i ≤ m. Geometricamente una somma di Riemann e` costituita dalla somma delle aree dei rettangoli di base gli intervalli [ti , ti+1 ] e altezza g(ti ). Quando gli intervalli della suddivisione diventano piccoli, una somma di

Figura 1.2. Significato di una somma di Riemann

Riemann di g si avvicina all’area del trapezoide di g . Pi`u precisamente:  b lim g(t0 )(t1 − t0 ) + ... + g(tm )(tm+1 − tm ) = g(t) dt. max{ti+1 −ti }→0

a

Esempio 1.8 Sia m naturale, m ≥ 2. Allora

1 + e1/m + e2/m + ... + e(m−1)/m = e − 1. m→+∞ m lim

Infatti

1 + e1/m + e2/m + ... + e(m−1)/m 1 1 1 = e0 + e1/m + ... + em/m m m m m e` una somma di Riemann di ex su [0, 1], relativa alla suddivisione m−1 m 1 < ... < < =1: 0< m m m  1 il limite vale quindi et dt = e − 1. 0

Esercizi Esercizio 1.1 Individuare le risposte corrette. Il grafico di una funzione continua h : [a, b] → R e` a) Una curva; b) Una curva parametrica; c) Il sostegno di una curva parametrica. Esercizio 1.2 La cardioide e` la curva di equazione polare r = a(1 + cos θ), θ ∈ [0, 2π], dove a > 0 e` una costante; calcolarne la lunghezza L.

Note sul Capitolo 1

7

Figura 1.3. Cardioide.

Soluzioni degli esercizi Soluzione es. 1.1. a) e c) Si, infatti si tratta del sostegno della curva x ∈ [a, b] → (x, h(x)) ∈ R2 . b) No, una curva parametrica e` una funzione, un insieme non e` mai una curva parametrica. Soluzione es. 1.2.  2  2π  d (a(1 + cos θ)) + (a(1 + cos θ))2 dθ L= dθ 0  2π  =a sin2 θ + 1 + 2 cos θ + cos2 θ dθ = 0  2π  =a 2(1 + cos θ) dθ 0   2π  2π  π 1 + cos θ = 2a dθ = 2a | cos(θ/2)| dθ = 4a cos(θ/2) dθ = 2 0 0 0 π = 8a [sin(θ/2)]0 = 8a.

2

Note sul Capitolo 2

Definizione di curva parametrica nello spazio (Sez. 2.1) Notazione. Nel Libro si indica con |x| la norma euclidea in Rn : se x = (x1 , x2 , ..., xn ) si ha  |x| = x21 + ... + x2n . Analogamente a quanto visto nel Capitolo 1, precisiamo la definizione di curva parametrica. Definizione 2.1 (Curva parametrica) Una curva parametrica in Rn e` una funzione vettoriale continua a valori in Rn r(t) = (f1 , f2 , .., fn ) : I → Rn ,

t → (f1 (t), f2 (t), ..., fn (t))

dove f1 , ..., fn sono funzioni continue definite su un intervallo non degenere I di R. L’insieme dei punti (x1 , ..., xn ) = (f1 (t), ..., fn (t)), al variare di t ∈ I , si chiama sostegno della curva parametrica, o semplicemente curva di cui (f1 , ..., fn ) e` una parametrizzazione.

 Osservazione 2.2 (Una curva parametrica non e` un insieme di punti) Una curva parametrica nello spazio non e` un insieme ma una terna di funzioni continue definite su un intervallo. Una curva parametrica parametrizza una ed una sola curva (l’insieme immagine della curva parametrica). Viceversa, una curva pu`o essere parametrizzata in infiniti modi.

3

Note sul Capitolo 3

Funzioni di piu` variabili (Sez. 3.1) Approfondiamo qui qualche concetto utile di topologia del piano o dello spazio. Se p ∈ Rn e r > 0 indicheremo con B(p, r[ (risp. B(p, r]) la palla aperta (risp. chiusa) definita da B(p, r[:= {x ∈ Rn : |x − p| < r},

B(p, r[:= {x ∈ Rn : |x − p| ≤ r}. Per n = 2 si parla di disco piuttosto che di palla. Definizione 3.1 (Intorno) Sia p ∈ Rn . Un intorno di p e` un insieme che contiene un disco (n = 2) o palla (n > 2) di centro p e raggio r > 0. Rinviamo al testo per la definizione di punto interno e di frontiera. Definizione 3.2 (Insieme aperto) Un insieme si dice aperto se ogni suo punto e` interno.

Limiti di funzioni di piu` variabili (Sez. 3.2 ) Nella definizione del testo e` sottinteso il fatto che la variabile appartiene al dominio della funzione in un punto di accumulazione. Vediamo di precisare meglio i concetti e l’ambito di validit`a di tale definizione. Il limite in un punto ha significato solo se il punto e` sufficientemente “vicino” al dominio della funzione. Definizione 3.3 (Punto di accumulazione) Sia D un sottoinsieme di Rn . p ∈ Rn e` di accumulazione per D se ogni intorno di p contiene punti di D diversi da p. Ecco la definizione precisa di limite per una funzione di pi`u variabili.

12

Note sul Capitolo 3

Definizione 3.4 (Limite) Siano D un sottoinsieme non vuoto di Rn , f : D → R una funzione e p ∈ Rn di accumulazione per D. Si dice che L ∈ R e` il limite di f (x) per x che tende a p, e si scrive x→p lim f (x) = L, se x∈D

∀ε > 0 ∃δ > 0 :

x ∈ D,

0 < |x − p| < δ ⇒ |f (x) − L| < ε.

Il fatto che p sia di accumulazione per D garantisce nella definizione di limite che esista almeno un x ∈ D tale che 0 < |x − p| < δ . Nella definizione di limite abbiamo evidenziato il ruolo del dominio. Negli esempi del Libro si utilizza il fatto che se il limite di f esiste, allora esiste ed e` uguale il limite di f su una qualunque restrizione. La semplice dimostrazione della proposizione che segue e` un utile esercizio sulla definizione di limite e viene lasciata al lettore. Proposizione 3.5 (Limite sulle restrizioni) Siano f : D ⊆ Rn → R e E ⊆ D, con p di accumulazione per E . Si supponga che x→p lim f (x) = L. Allora x∈D

lim f (x) = L. x→p x∈E

Esempio 3.6 (Le coordinate polari nel piano) Nel piano, un metodo alternativo a quello delle restrizioni sulle rette e` quello delle coordinate polari. Siano p ∈ R2 f : B(p, r]\{p} → R una funzione. Nel calcolare il limite di f in p = (p1 , p2 ) ∈ R2 si pu`o porre

x = p1 + ρ cos t, y = p2 + ρ sin t ρ > 0, t ∈ [0, 2π], e studiare, per t fissato, il limite della composta

lim f (p1 + ρ cos t, p2 + ρ sin t).

ρ→0+

Ci`o equivale a studiare il limite di f sulla semiretta per p che fa un angolo t rispetto xy all’asse orientato x. Ad esempio, sia f (x, y) = 2 per (x, y) = (0, 0): studiax + y2 mone il limite in (0, 0). Posto x = ρ cos t, y = ρ sin t, con ρ > 0 e t ∈ [0, 2π] si ha f (ρ cos t, ρ sin t) = cos t sin t −→ρ→0+ cos t sin t. tale limite dipende da t: si conclude che il limite di f in (0, 0) non esiste. Di conseguenza, se su due restrizioni si trovano limiti diversi, il limite non esi-

 ste. Si faccia anche attenzione che il limite di una funzione in p pu`o esistere ed essere uguale su ogni retta per il punto p, senza che il limite della funzione esista. analogamente, pu`o benissimo essere che

∀t ∈ [0, 2π]

lim f (p1 + ρ cos t, p2 + ρ sin t) = L,

ρ→0+

Note sul Capitolo 3

13

ma che il limite di f in p = (p1 , p2 ) non esista.

2x2 y definita per (x, y) = (0, 0) non ha x4 + y 2 limite (si veda l’Esempio 6 del Libro). Tuttavia, Esempio 3.7 La funzione f (x, y) =

∀t ∈ [0, 2π] f (ρ cos t, ρ sin t) = ρ

2 cos2 t sin t →0 ρ2 cos4 t + sin2 t

per ρ → 0+ : verificarlo separatamente per t ∈ {0, π, 2π} (la funzione composta vale 0) e poi per t ∈ / {0, π, 2π} (il denominatore tende a sin2 t = 0). Tuttavia, se il limite di f in p esiste e vale L su un ricoprimento finito del dominio allora il limite di f in p vale L. Proposizione 3.8 Siano D un sottoinsieme non vuoto di Rn , f : D → R una funzione e p ∈ Rn di accumulazione per D. Si supponga che D = D1 ∪ D2 ∪ ... ∪ Dm e che per ogni i = 1, ..., m, x→p lim f (x) = L. Allora x→p lim f (x) = L. x∈Di

x∈D

La dimostrazione della proposizione e` un facile esercizio (consigliato). '๛ '๜

'๟ S

'๝

'๞

Figura 3.1. Un ricoprimento finito del dominio

Esempio 3.9 Sia

x2 y f (x, y) = sin(xy)

x ≤ y ≤ 2x . altrimenti

Per determinare il limite in (0, 0) poniamo D1 := {(x, y) : x ≤ y ≤ 2x}, D2 := R2 \ D1 . Per la continuit`a delle funzioni x2 y e sin(xy) si ha

lim

(x,y)→(0,0) (x,y)∈D1

lim

(x,y)→(0,0) (x,y)∈D2

f (x, y) =

f (x, y) =

lim

(x,y)→(0,0) (x,y)∈D1

lim

(x,y)→(0,0) (x,y)∈D2

x2 y = 02 × 0 = 0,

sin(xy) = sin(0) = 0.

Per la Proposizione 3.8 si conclude che il limite di f in (0, 0) vale 0.

14

Note sul Capitolo 3

3.0.1 Metodi per provare l’esistenza di un limite Provare che un limite di una funzione non esiste in un punto e` spesso pi`u facile che provarne l’esistenza: e` sufficiente trovare due restrizioni per quel punto lungo le quali la funzione ha un limite diverso. Il problema nasce quando, provando su varie restrizioni, il limite di una funzione in un punto e` sempre lo stesso. In tal caso le strade sono due: cercare una restrizione pi`u ingegnosa lungo la quale la funzione ha un comportamento diverso, o provare l’esistenza del limite intuito in precedenza. Usare i limiti notevoli e le asintoticit`a Tutti i risultati visti sulle asintoticit`a e loro uso nei limiti continua a valere per funzioni di pi`u variabili. Ci limitiamo ad un esempio Studiamo l’esistenza del limite

2 sin(x2 y) . (x,y)→(0,0) x4 + y 2 lim

Dato che x2 y tende a 0 per (x, y) → (0, 0) e che sin t ∼ t per t → 0 si ha che il limite cercato vale 2x2 y lim , (x,y)→(0,0) x4 + y 2 che non esiste, come si e` visto nell’Esempio 3.7. Stime uniformi Quando si intuisce il valore del limite (ad esempio scrutando il limite su varie restrizioni) e` spesso utile il seguente criterio di confronto. Proposizione 3.10 (Stima uniforme) Siano D un sottoinsieme non vuoto di Rn , f : D → R una funzione e p ∈ Rn di accumulazione per D. Sia L ∈ R e si supponga

∀x ∈ D \ {p} |f (x) − L| ≤ h(x),

lim h(x) = 0.

x→p

Allora lim f (x) = L. x→p

Dimostrazione. Si ha L − h(x) ≤ f (x) ≤ L + h(x): si conclude per il Teorema dei Carabinieri.  Esempio 3.11 Sia f (x, y) =

x2 y , (x, y) = (0, 0). Si noti che vale la seguente x2 + y 2

Disuguaglianza importante:

1 |xy| ≤ (x2 + y 2 ) 2

che discende semplicemente dal fatto che (|x| − |y|)2 ≥ 0. Pertanto

1 |f (x, y)| ≤ |x| → 0 per (x, y) → (0, 0) : 2 per la Proposizione 3.10 si conclude che il limite cercato vale 0.

Note sul Capitolo 3

15

Nel piano, abbiamo visto nell’Esempio 3.6 che il cambio in polari e` un utile strumento per “indovinare” il limite di una funzione nel centro delle coordinate. Ecco una stima uniforme rispetto agli angoli che permette di provare che il limite e` quello intuito facendo i limiti per il raggio ρ → 0+ per l’angolo t fissato. Esempio 3.12 (Limiti con le coordinate polari) Siano p ∈ R2 f : B(p, r] \ {p} → R una funzione. Esista h :]0, r] → [0, +∞[ tale che

∀ρ ∈]0, r] |f (p1 + ρ cos t, p2 + ρ sin t) − L| ≤ h(ρ), con lim+ h(ρ) = 0. Allora ρ→0

lim

(x,y)→(0,0)

(3.1)

f (x, y) = L. Infatti, si fissi ε > 0, e sia δ > 0

tale che

0 < ρ < δ ⇒ h(ρ) < ε. Allora segue da (3.1) che

0 < |(x, y) − (p1 , p2 )| < δ ⇒ |f (x, y) − L| < ε, da cui la conclusione. Ad esempio, si voglia calcolare il limite visto nell’Esempio 3.11. Per ogni ρ > 0, t ∈ [0, 2π] si ha f (ρ cos t, ρ sin t) = 2ρ cos2 t sin t → 0 per ρ → 0+ . Come abbiamo visto sopra, ci`o non e` sufficiente per concludere sull’esistenza del limite (ma solo che se esso esiste, vale 0). Tuttavia vale la seguente stima uniforme, indipendente da t:

|f (ρ cos t, ρ sin t)| = 2ρ| cos2 t sin t| ≤ h(ρ) := 2ρ ed e` lim+ h(ρ) = 0: per quanto visto sopra ci`o basta per concludere che il limite ρ→0

cercato vale 0.

Derivate parziali (Sez. 3.3) Notazione. La derivata parziale di una funzione f rispetto ad xi verr`a anche indicata  con ∂xi f (x). Indicheremo con la notazione standard Du f (p), al posto che df usata nel Libro, la derivata direzionale di una funzione f nel punto p, con ds u,p rispetto al vettore u.

3.0.2 Derivate parziali Come per le derivate di funzioni di una variabile, non e` detto che le derivate parziali siano continue, cio`e che se f : B(p, r[⊂ R2 → R si abbia

∂xi f (p) = lim ∂xi f (x). x→p

16

Note sul Capitolo 3

Esempio 3.13 Sia

f (x, y) = Si ha

⎧ ⎨

x2

⎩0

xy + y2

se (x, y) = (0, 0) se (x, y) = (0, 0).

f (t, 0) − f (0, 0) 0 = lim = 0. t→0 t→0 t t

∂x f (0, 0) = lim Per (x, y) = (0, 0) si ha

∂x f (x, y) =

y (y 2 − x2 ) 2 . (x2 + y 2 )

Il limite

lim

(x,y)→(0,0)

non esiste dato che

∂x f (x, y)

y (y 2 − x2 ) 2 = 0, (x,y)→(0,0) (x2 + y 2 ) x=y lim

y (y 2 − x2 ) 2 = +∞. (x,y)→(0,0) (x2 + y 2 ) x=0,y>0 lim

3.0.3 Differenziabilita`  Nella definizione di differenziabilit`a data nel Libro, ε1 , ε2 non sono ”valori” bens`ı funzioni. Pi`u precisamente si ha: Definizione 3.14 Se R e` aperto in R2 , f : R → R e (x0 , y0 ) ∈ R, si dice che f e` differenziabile in (x0 , y0 ) se esistono fx (x0 , y0 ) e fy (x0 , y0 ) ed inoltre

f (x0 + Δx, y0 + Δy) − f (x0 , y0 ) = = fx (x0 , y0 )Δx + fy (x0 , y0 )Δy + ε1 (x, y)Δx + ε2 (x, y)Δy, (3.2) con ε1 (x, y), ε2 (x, y) → 0 per (x, y) → (x0 , y0 ). Osservazione 3.15 Un modo equivalente per scrivere (3.2) e`

f (x, y) = f (x0 , y0 ) + ∂x f (x0 , y0 )(x − x0 )+ + ∂y f (x0 , y0 )(y − y0 ) + o(|(x − x0 , y − y0 )|) = f (x0 , y0 ) + ∇f (x0 , y0 ) · (x − x0 , y − y0 ) + o(|(x − x0 , y − y0 )|), (3.3)

Note sul Capitolo 3

17

dove con o(|(x − x0 , y − y0 )|) indichiamo, similmente a quanto visto in Analisi Uno, una funzione R(x, y) tale che

lim

(x,y)→(x0 ,y0 )



R(x, y) = 0. (x − x0 )2 + (y − y0 )2

Il Teorema 8 del Libro mostra che se f e` differenziabile in un punto p allora Du f (p) = ∇f (p) · u. Pertanto, in tal caso la derivata direzionale e` una funzione lineare in u, cio`e del tipo u = (u1 , u2 ) → au1 + bu2 per qualche a, b reali. Esempio 3.16 Sia f : R2 → R definita da ⎧ ⎨

x3 f (x, y) = x2 + y 2 ⎩ 0

se (x, y) = (0, 0) se (x, y) = (0, 0).

1. f e` continua in (0, 0). Infatti |f (x, y)| ≤ |x| x2 /(x2 +y 2 ) ≤ |x| e

0 quindi

lim

(x,y)→(0,0)

f (x, y) = (0, 0) = f (0, 0).

lim

(x,y)→(0,0)

|x| =

u = (0, 0). Si ha 2. Sia u = (u1 , u2 ) ∈ R2 ,  f (tu) − f (0) f (tu1 , tu2 ) = lim t→0 t t t3 u31 u31 . = 2 = lim 3 2 t→0 t (u + u2 ) u1 + u22 1 2

Du f (0, 0) = lim t→0

3. f non e` differenziabile in (0, 0). Infatti l’applicazione u → Du f (0, 0) non e` lineare, ad esempio e` D(1,1) f (0, 0) = 1/2 mentre D(1,0) f (0, 0) + D(0,1) f (0, 0) = 1 + 0 = 1.

Figura 3.2. Una funzione non differenziabile: si vede come il piano tangente non

fornisce una buona approssimazione dei valori della funzione attorno al punto considerato

18

Note sul Capitolo 3

Derivate direzionali e vettori gradiente (Sez. 3.5) Notazione. Indicheremo con ∇f (p) il gradiente di una funzione f in un punto p. La regola della catena vista nella Sezione 3.4 si pu`o sinteticamente enunciare in termini del gradiente. Proposizione 3.17 (Regola della catena) Se f : R ⊂ Rn → R e` differenziabile e r : [a, b] → R e` derivabile allora

d (f ◦ r)(t) = ∇f (r(t)) · r (t). dt E’ spesso utile ricordare il gradiente della norma euclidea.

 Esempio 3.18 (Gradiente della norma) Per ogni x = 0 si ha ∇|x| =

x . |x|

Infatti, per ogni i = 1, ..., n si ha  ∂xi |x| = ∂xi x21 + ... + x2n = 

xi xi . = |x| + ... + x2n √ Si noti che |x| non ha derivate parziali nell’origine, dato che t → t2 = |t| non e` derivabile on 0. x21

Valori estremi e punti di sella (Sez. 3.7) Il criterio delle derivate seconde illustrato nel Libro permette di concludere la natura di un punto critico per una funzione f in dimensione 2 non appena f ha derivate 2 seconde continue (cio`e e` di classe C 2 ) e fxx fyy − fxy = 0. Negli altri casi bisogna procedere diversamente. Infatti quando il determinante della matrice Hessiana e` nullo pu`o accadere di tutto. Esempio 3.19 (Determinante Hessiano nullo) • f (x, y) = x4 +y 4 ha (0, 0) come punto critico, il determinante Hessiano di f in (0, 0) e` nullo. Si vede che (0, 0) e` minimo (globale) dato che f (x, y) ≥ 0 = f (0, 0) per ogni (x, y) ∈ R2 . • f (x, y) = −(x4 + y 4 ) ha (0, 0) come punto critico, il determinante Hessiano di f in (0, 0) e` nullo. Si vede che (0, 0) e` massimo (globale) dato che f (x, y) ≤ 0 = f (0, 0) per ogni (x, y) ∈ R2 . • f (x, y) = x4 − y 4 ha (0, 0) come punto critico, il determinante Hessiano di f in (0, 0) e` nullo. Si vede che (0, 0) e` di sella dato che f (x, 0) > 0 = f (0, 0) e f (0, y) < 0 = f (0, 0) per ogni (x, y) ∈ R2 con x = 0, y = 0.

Note sul Capitolo 3



19

E’ importante notare che la definizione di minimi/massimo locale non coinvolge nessuna derivata della funzione: si tratta di verificare che in un intorno B(p, r[ di un punto p fissato vale una disuguaglianza di questo tipo:

f (x) − f (p) ≥ 0 ∀x ∈ U oppure

f (x) − f (p) ≤ 0 ∀x ∈ U. Esempio 3.20 (Un punto di sella che e` minimo locale su ogni retta che lo contiene) Sia f (x, y) = y 2 − 3x2 y + 2x4 . Studiamone punti critici e natura. Si ha ∇f (x, y) = (−6xy + 8x3 , 2y − 3x2 ). Pertanto (x, y) e` critico se e solo se   −6xy + 8x3 = 0 2x(4x2 − 3y) = 0 ⇔ 2y − 3x2 = 0 2y − 3x2 = 0 da cui segue subito x = y = 0. La matrice Hessiana in (x, y) vale  24x2 − 6y −6x H(x, y) = −6x 2  0 0 , il cui determinante e` nullo: il criterio delIn (0, 0) si ha H(0, 0) = 0 2 l’Hessiano non fornisce informazioni. Studiamo il segno di f (x, y) − f (0, 0) = y 2 −3x2 y +2x4 attorno a (0, 0). Sembra difficile immaginare che l’espressione scritta abbia un segno costante in un disco centrato nell’origine, vediamo di dimostrarlo. Supponiamo ad esempio che sia

∀(x, y) ∈ B((0, 0), r[ y 2 − 3x2 y + 2x4 ≥ 0. Allora y 2 + 2x4 ≥ 3x2 y attorno all’origine. Consideriamo i punti della parabola y = ax2 , con a ∈ R. La disuguaglianza diventa (a2 + 2) ≥ 3a, che e` soddisfatta solo per a ≤ 1 o a ≥ 2. Pertanto essa e` violata ad esempio sui punti del tipo y = 3 2 x . Il ragionamento appena svolto mostra che f (x, y) − f (0, 0) < 0 se e solo se 2 x2 < y ≤ 2x2 , ed e` f (x, y) − f (0, 0) > 0 se e solo se y < x2 o y > 2x2 : si poteva giungere a questa conclusione osservando che f (x, y) = (y − x2 )(y − 2x2 ). Pertanto (0, 0) e` di sella. Si noti che (0, 0) e` minimo locale per f su ogni retta per l’origine. Infatti:

• Sulla retta x = 0 si ha f (x, y) − f (0, 0) = y 2 ≥ 0, • Sulla retta y = 0 si ha f (x, y) − f (0, 0) = 2x4 ≥ 0, • Su una generica altra retta per l’origine y = bx con b ∈ R, b = 0 si ha f (x, y) − f (0, 0) = x2 (b − x)(b − 2x) ≥ 0 |b| per |x| < √ . 2

20

Note sul Capitolo 3























Figura 3.3. La regione dove f (x, y) − f (0, 0) < 0

Esercizi Esercizio 3.1 Sia f → R2 \ {(0, 0)} → R definita da ⎧ 2 ⎨ xy se (x, y) = (0, 0); f (x, y) = x2 + y 2 ⎩ 0 se (x, y) = (0, 0). (i) Studiare la continuit`a di f in (0, 0). (ii) Dire se f ammette le derivate parziali ∂x f (0, 0) e ∂y f (0, 0) in (0, 0); calcolarle in caso affermativo. (iii) Dire se f e` differenziabile in (0, 0). Determinare df (0, 0) in caso affermativo. Esercizio 3.2 Sia f : R2 → R di classe C 1 . Si supponga che f (et , t) = t2 e che f (t2 + 1, sin t) = t. Determinare ∇f (1, 0).

a)(−1, 1)

b)(1, −2)

c)(2, 1)

d)(1, −1)

e)(−1, −1)

Esercizio 3.3 Trovare massimo e minimo assoluti e i punti di massimo e minimo assoluti della funzione

f (x, y) = x2 − 4x + arctan(4y + y 2 ) ristretta al triangolo T di vertici O = (0, 0), A = (0, −4), B(4, 0).

Soluzioni degli esercizi Soluzione es. 3.1. (i) Ricordando che |xy| ≤ 12 (x2 + y 2 ) si ha 1 (x2 + y 2 ) |xy| 2 ≤ |y| x2 + y 2 x2 + y 2 1 2 1 ≤ |y| ≤ x + y 2 −−−−−−−→ 0 (x,y)→(0,0) 2 2

|f (x, y)| ≤ |y|

Note sul Capitolo 3

da cui

lim

(x,y)→(0,0)

21

f (x, y) = 0 = f (0, 0). Quindi f e` continua nell’origine.

(ii) Per definizione di f , si ha che f e` identicamente nulla sugli assi cartesiani. Quindi le derivate parziali esistono nell’origine e valgono

∂f ∂f (0, 0) = 0 = (0, 0). ∂x ∂y (iii) Per il punto (ii), se f e` differenziabile nell’origine allora il suo differenziale e` la funzione nulla. Si ha

∂f ∂f (0, 0)x − (0, 0)y xy 2 ∂x ∂x √ = 2 . |(x, y)| (x + y 2 ) x2 + y 2

f (x, y) − f (0, 0) −

Posto x = ρ cos t, y = ρ sin t (ρ ≥ 0, t ∈ [0, 2π]) e`

xy 2 √ = cos t sin2 t, (x2 + y 2 ) x2 + y 2

pertanto

xy 2 √ (x,y)→(0,0) (x2 + y 2 ) x2 + y 2 lim

non esiste ed f non e` differenziabile in (0, 0). Soluzione es. 3.2. d) Soluzione es. 3.3. E` 

 1 ∇f (x, y) = 2x − 4, (4 + 2y) . 1 + (4y + y 2 )2 L’unico punto critico e` (2, −2), che non e` interno a T : minimo e massimo assoluti di f vanno cercati sulla frontiera ∂T di T .

y O

2

B x

T −2

A Figura 3.4.

22

Note sul Capitolo 3

0 2 4 x f1(x) − − − − − − − − 0 + + + + + + + 0 f1(x)

0 −4

Figura 3.5.

Su OB e` y = 0, f (x, 0) = x2 − 4x = f1 (x), f1 (x) = 2x − 4 e maxOB f = f1 (0) = 0, minOB f = f1 (2) = −4 (Figura 3.5). Su OA e` x = 0, f (0, y) = f2 (y) = arctan(4y + y 2 ). Posto g(y) = y 2 + 4y e` g  (y) = 2y + 4; arctan e` crescente quindi minOA f = arctan(−4), maxOA f = arctan 0 = 0. 0 −4 −2 y g (y) − − − − − − − − 0 + + + + + + + 0

0 g(y)

−4

Figura 3.6.

Su AB e` y = x − 4 e f (x, x − 4) = x2 − 4x + arctan(x2 − 4x) = u + arctan u = h(u), con u = x2 − 4x. Quando x descrive l’intervallo [0, 4], u varia tra −4 e 0. E` h (u) = 1 + 1/(1 + u2 ) > 0 quindi h e` strettamente crescente e minx∈[0,4] f (x, x − 4) = h(−4) = −4 + arctan(−4) = minAB f = f (2, −2), maxx∈[0,4] f (x, x − 4) = h(0) = 0 = maxAB f . Pertanto minT f = −4 + arctan(−4) = f (2, −2); maxT f = 0 = f (0, 0) = f (A) = f (B).

4

Note sul Capitolo 4

Integrali tripli in coordinate rettangolari (Sez. 4.5) Nel Libro viene illustrato il metodo di integrazione su domini semplici: si fanno variare due coordinate, si integra prima rispetto alla terza. Vediamo qui un metodo di integrazione per “fette parallele agli assi coordinati”: si fa variare una coordinata e si integra prima rispetto a due coordinate. Vediamo il corrispondente matematico di una sottile fetta di polenta sul piatto. La z -sezione di D e` la proiezione sul piano xy della fetta parallela al piano xy , a quota z , dell’insieme D. Definizione 4.1 Sia D ⊆ R3 .

• La z -sezione di D e` il sottoinsieme di R2 definito da D(z) = {(x, y) ∈ R2 : (x, y, z) ∈ D}. • La proiezione πz (D) di D sull’asse z e` l’insieme degli z ∈ R tali che D(z) = ∅. Osservazione 4.2 Sia D = {(x, y, z) : g(x, y, z) ≤ 0}.

• Per z ∈ R la z -sezione di D e` D(z) = {(x, y) ∈ R2 : g(x, y, z) ≤ 0.} • La proiezione di D sull’asse z e` l’insieme {z ∈ R : ∃(x, y) ∈ R2 , g(x, y, z) ≤ 0}. Esempio 4.3 Sia D la palla chiusa di raggio 1,

D = {(x, y, z) : x2 + y 2 + z 2 ≤ 1} La z -sezione di D e` l’insieme

{(x, y) : x2 + y 2 ≤ 1 − z 2 }, questa e` non vuota se e solo se z ∈ [−1, 1]: πz (D) = [−1, 1].

24

Note sul Capitolo 4

Enunciamo la formula di integrazione per fette solo per fette parallele al piano xy , formule simili valgono sugli altri piani coordinati. Proposizione 4.4 (Formula di integrazione per fette) Sia f (x, y, z) funzione continua su una regione chiusa e limitata D di R3 . Allora     f (x, y, z) dx dy dz = f (x, y, z) dx dy dz. D

πz (D)

D(z)

Per quanto ci riguarda la applicazione della formula precedente si lmiter`a ai casi nei quali πz (D) e` unione finita di intervalli o D(z) e` unione di domini semplici. Nel caso di f = 1 la formula dice che il volume di un solido si calcola integrando l’area delle fette parallele ad un piano coordinato. Corollario 4.5 (Principio di Bonaventura-Cavalieri) Sia D chiuso e limitato di R3 . Allora  V ol(D) = Area (D(z)) dz. πz (D)

In particolare solidi con stesse aree delle sezioni hanno uguale volume. Esempio 4.6 (Volume della palla di raggio R) Calcoliamo il volume della palla chiusa BR di raggio R. Si ha che πz (BR ) = [−R, R], mentre per |z| ≤ R,

BR (z) = {(x, y) ∈ R2 : x2 + y 2 ≤ R2 − z 2 } √ e` il disco di raggio R2 − z 2 . La formula di integrazione per fette porge quindi 

V ol(BR ) =

R

−R R



=

−R

Area B(0, 2

√

R2 − z 2 ] dz

2

π(R − z ) dz = 2π

 0

R

(R2 − z 2 ) dz

  1 3 R 4 3 2 = 2π R z − z = πR . 3 3 0 Esempio 4.7 Si vuole calcolare il volume del solido

D = {(x, y, z) ∈ R3 : x2 + z 2 ≤ 1, y 2 + z 2 ≤ 1}. Si tratta dell’intersezione del cilindro di asse y dato da x2 + z 2 ≤ 1 con il cilindro di asse x definito da y 2 + z 2 ≤ 1. Il dominio e` normale rispetto al piano xy , infatti

D = {(x, y, z) ∈ R3 : −h(x, y) ≤ z ≤ h(x, y)},

Note sul Capitolo 4

25

Figura 4.1. L’insieme D

dove h(x, y) := min{1 − x2 , 1 − y 2 }. La formula compatta di h varia a seconda che sia |x| ≤ |y| o viceversa: e` pi`u agevole integrare per fette parallele al pano xy . Si ha πz (D) = [−1, 1]. Per |z| ≤ 1 si ha

D(z) = {(x, y) ∈ R2 : x2 ≤ 1 − z 2 , y 2 ≤ 1 − z 2 } √ √ √ √ = [− 1 − z 2 , 1 − z 2 ] × [− 1 − z 2 , 1 − z 2 ]. La formula di integrazione per fette parallele al piano yz porge allora  1  1 V ol(D) = Area D(z) dz = 4(1 − z 2 ) dz −1



=8 z−

3

z 3

−1

1 0

=

16 . 3

Cambio di variabili, solidi di rotazione e simmetrie(Sez. 4.8) 4.0.1 Cambio di variabile Il cambio di variabile in coordinate cilindriche e` utile quando la funzione da integrare dipende solo dalla distanza √ 2 del2 punto dall’asse z e dalla quota z . Pi`u precisamente se f (x, y, z) = g( x + y , z) allora il cambiamento di variabili in coordinate cilindriche porge   f (x, y, z) dx dy dz = rg(r, z) dr dt dz D

(r,t,z):(r cos t,r sin t,z)∈D

Una applicazione importante e` rappresentata dai solidi di rotazione.

Note sul Capitolo 4

Vol

z (E2π )

27



=

E2π

1 dx dy dz



= 2π



ρ dρ dz = 2π

(ρ,z): (ρ cos t,ρ sin t,z)∈E2π

ρ dρ dz. (ρ,z)∈E

 L’enunciato del Teorema di Pappo - Guldino per i solidi di rotazione fa intervenire due formule equivalenti: si usa l’una o l’altra a seconda delle simmetrie e facilit`a del calcolo dell’area dell’insieme piano E . Esempio 4.10 (Volume del toro) Un toro (dal latino torus, anello, mentre lanimale e` taurus) e` il solido che si ottiene facendo ruotare (di 2π ) un disco di raggio a > 0 attorno ad una retta che dista R > a dal centro del disco (una ciambella, od una camera daria di pneumatico pensata piena realizzano un toro). Chiaramente il baricentro del disco e` il centro del disco. Ne segue che il volume del toro e` 2πR(πa2 ) = 2π 2 Ra2 . Esempio 4.11 Vogliamo calcolare il volume del solido ottenuto ruotando il rettangolo E = [2, 8] × [6, 8] attorno all’asse z . L’area di E vale 12, l’ascissa del baricentro vale 5: il volume cercato vale quindi 2π × 12 × 5 = 120π . La definizione e il risultato per solidi di rotazione attorno all’asse x o y sono del

 tutto analoghe a quelle date sopra. Ricordarsi che interviene la distanza del baricentro dall’asse di rotazione, indipendentemente dal nome dell’asse. Esempio 4.12 Vogliamo calcolare il volume del solido ottenuto ruotando il rettangolo E = [2, 8] × [6, 8] attorno all’asse x. L’area di E vale 12, l’ordinata del baricentro (che e` la distanza dall’asse di rotazione) vale 7: il volume cercato vale quindi 2π × 12 × 7 = 168π . x ottenuto ruotando di Esempio 4.13 (Metodo dei dischi) Si considera il solido E2π 2π attorno all’asse x l’insieme E del piano xz con z ≥ 0 delimitato dal grafico di due funzioni f e g positive:

E = {(x, z) : x ∈ [a, b], f (x) ≤ z ≤ g(x)}. Si ha Vol

x (E2π )



=

a

b

πg(x)2 − πf (x)2 dx.

Possiamo provarlo in due modi: 1. Utilizzando il metodo di integrazione per fette parallele al piano yz (Proposizione 4.4) si ottiene subito  b  b x x Area (E2π (x)) dx = πg(x)2 − πf (x)2 dx. Vol (E2π ) = a

a

28

Note sul Capitolo 4

2. Utilizzando la formula di Pappo - Guldino si ha  x Vol (E2π ) = 2π z dx dz E b



= 2π

a



= 2π

a



g(x)

z dz dx

f (x)

b



1 (g(x)2 − f (x)2 2





dx =

a

b

πg(x)2 − πf (x)2 dx.

4.0.2 Simmetrie Il calcolo di integrali e` talvolta facilitato dalla presenza di simmetrie. Vediamo un esempio. Esempio 4.14 Sia D ⊂ R2 simmetrico rispetto all’asse y e f : D → R tale che  f (−x, y) = −f (x, y) per ogni (x, y) ∈ D. Allora f (x, y) dx dy = 0. Infatti il D

cambio di variabile u = −x, v = y porge    f (x, y) dx dy = f (−u, v) du dv = − f (u, v) du dv D D D  f (x, y) dx dy, =− D

da cui l’asserto. Cos`ı ad esempio, se D = {(x, y) ∈ R2 : x2 + y 2 ≤ 1}, l’integrale  sin(x3 y 8 ) log(1 + x2 y 2 ) dx dy = 0. D

In modo analogo si possono giustificare varie formule di simmetria; ne citiamo uno tra i vari che si possono presentare. Esempio 4.15 Sia D ⊂ R3 simmetrico rispetto all’origine e f : D → R  tale che f (−x, −y, −z) = −f (x, y, z) per ogni (x, y, z) ∈ D. Allora D

f (x, y, z) dx dy dz = 0.

Esercizi Esercizio 4.1 Determinare il volume del solido compreso tra i piani x = 0 e x = 1, le cui sezioni perpendicolari all’asse√x sono dischi circolari i cui diametri vanno dalla parabola y = x2 alla parabola y = x. Esercizio 4.2 Determinare il volume del solido generato dalla rotazione attorno agli assi dati qui sotto della regione delimitata dall’asse x, dalla curva y = 3x4 e dalle rette x = 1 e x = −1:

Note sul Capitolo 4

29

(a) Rotazione attorno all’asse x; (b) Rotazione attorno all’asse y . Esercizio 4.3 Determinare il volume del solido generato dalla rotazione attorno agli assi dati qui sotto della regione delimitata a sinistra dalla parabola x = y 2 + 1 e a destra dalla retta x = 5: (a) Rotazione attorno all’asse x; (b) Rotazione attorno all’asse y . Esercizio 4.4 Determinare il volume del solido generato dalla rotazione attorno alπ l’asse x della regione delimitata dall’asse x, dalla retta x = e dalla curva 3 y = tan x nel primo quadrante del piano xy . Esercizio 4.5 Determinare il volume del solido generato dalla rotazione attorno al2 l’asse x della regione delimitata dalla curva x = ey e dalle rette y = 0, x = 0 e y = 1.

Soluzioni degli esercizi Soluzione es. 4.1.

9π . 280

Soluzione es. 4.2. (a) 2π

(b) π .

Soluzione es. 4.3. (a) 8π

(b)

1088π . 15

√ π(3 3 − π) Soluzione es. 4.4. . 3 Soluzione es. 4.5. π(e − 1).

5

Note sul Capitolo 5

Definizione di integrali di linea (Sez. 5.1 e Sez. 5.2) Nel Testo, se r(t), t ∈ [a, b] e` una curva parametrica C 1 con sostegno C , e f : C → R e` funzione, si indica con  f (x) ds (x = (x1 , ..., xn )) C

l’integrale curvilineo di f , dato da  a

b

f (r(t))|r |(t) dt.

 La notazione e` vagamente abusiva, dato che tale integrale dipende dalla parametrizzazione della curva: in altri testi esso viene infatti indicato con maggior precisione con  f (x, y, z) ds. r

Esempio 5.1 Si consideri il cerchio unitario C con le due parametrizzazioni: r1 (t) = (cos t, sin t), t ∈ [0, 2π] r2 (t) = (cos t, sin t), t ∈ [0, 4π] Si ha





1 ds = 2π, r1

1 ds = 4π. r2

Analogamente, se F : C → Rn e` un campo vettoriale definito su C preferiamo le notazioni   F · T ds, F · dr r

r

alle notazioni del Testo nella quale compare C . E’ spesso comodo integrare su curve C 1 a tratti.

32

Note sul Capitolo 5

Definizione 5.2 (Integrali su curve C 1 a tratti) Una curva r : [a, b] → Rn e` C 1 a tratti se r e` continua ed esistono t0 = a < t1 < ... < tm = b tali che r e` C 1 su ognuno degli intervalli [ti , ti+1 ]. L’integrale di un campo (scalare o vettoriale) su r e` in tal caso definito come la somma degli integrali del campo sulle curve ottenute restringendo r sugli intervalli [ti , ti+1 ]. Tipicamente si ottiene un cammino C 1 a tratti giustapponendo due cammini C 1 . Esempio 5.3 (Giustapposizione di due cammini) Siano r1 : [a, b] → Rn e r2 : [c, d] → Rn tali che il termine r1 (b) di r coincida con l’inizio r2 (c) di γ . La giustapposizione di r1 con r2 e` il cammino r1 r2 ottenuto percorrendo prima r1 e poi r2 , definito da  r1 (t) se t ∈ [a, b] ∀t ∈ [a, b + (d − c)] r1 r2 (t) = r2 (t + c − b) se t ∈ [b, b + (d − c)]. Se r1 e r2 sono C 1 allora r1 r2 e` C 1 a tratti. Inoltre, se f e` campo scalare definito sul

Figura 5.1. La giustapposizione di r1 , r2 , r3 , r4 .

sostegno di r1 r2 si ha 





f (x) ds = r1 r2

f (x) ds + r1

f (x) ds. r2

Una formula analoga vale per integrali di campi vettoriali.

Note sul Capitolo 5

33

Esempio 5.4 (Cammino inverso) Spesso una curva C 1 a tratti viene detta anche cammino. Sia r : [a, b] → Rn una curva parametrica. Il cammino inverso di r, indicato da r−1 e` definito da r−1 : [a, b] → Rn ,

r−1 (t) := r(b − (t − a)) ∀t ∈ [a, b].

Si noti che b − (t − a) ∈ [a, b] per ogni t ∈ [a, b], che r−1 e r hanno stesso sostegno, percorso in verso opposto: r−1 (a) = r(b), r−1 (b) = r(a). E’ ben noto che la massa di un filo metallico non cambia se si percorre il filo al contrario, mentre il lavoro di una forza diventa opposto svolgendo il percorso all’inverso. Pi`u precisamente, siano f : C → R un campo scalare e F : C → Rn un campo vettoriale continui. Il lettore verifichi usando la definizione che   f (x) ds = f (x) ds 

r−1

r−1

r



F · T ds = −

F · T ds. r

Campi conservativi (Sez. 5.3) Errata corrige: nella Sez. 5.3, prima della definizione di potenziale, sostituire “ipotesi di derivabilit`a” con “ipotesi di continuit`a”. Osservazione 5.5 Nel Libro viene chiamato potenziale di F una funzione scalare f tale che ∇f = F. In matematica si preferisce spesso il termine alternativo di primitiva, dato che per i fisici generalmente un potenziale e` un campo scalare u tale che ∇u = −F. Una classe notevole di campi gradiente (e quindi conservativi sugli aperti connessi) e` rappresentata dai campi radiali, tanto frequenti in Fisica. Definizione 5.6 Sia D ⊂ R3 un aperto del tipo

D = {x ∈ Rn : a < |x| < b ≤ +∞} (anello) o una palla aperta (eventualmente tutto Rn ). Un campo F : D → Rn si dice radiale se esiste una funzione continua di variabile reale ϕ tale che F(x) = ϕ(|x|)

x |x|

∀x ∈ D \ {0}.

Pertanto l’intensit`a ϕ(|x|) del campo e` uguale su ogni sfera contenuta in D, e il campo in un punto x ha la stessa direzione di x (eventualmente verso opposto a seconda del segno di ϕ).

34

Note sul Capitolo 5

Esempio 5.7 Il campo gravitazionale in R3 generato da una massa nell’origine`e del tipo (x1 , x2 , x3 ) K K x  F(x, y, z) = = 2 . 2 2 2 |x| |x| x1 + x2 + x3 x21 + x22 + x23 per qualche costante K . L’intensit`a del campo gravitazionale e` infatti inversamente proporzionale al quadrato della distanza dalla massa che lo genera.

Proposizione 5.8 Ogni campo radiale e` un campo gradiente. Se F(x) = x ϕ(|x|) un potenziale di F e` dato da f (x) = u(|x|), dove u e` una primitiva di |x| ϕ. Dimostrazione. Infatti dalla regola della catena per ogni x = 0 si ha

∇u(|x|) = u (|x|)

x x = ϕ(|x|) = F(x). |x| |x| 

Esempio 5.9 Determiniamo una primitiva del campo gravitazionale dell’Esemx K pio 5.7. Il campo e` dato da F(x, y, z) = ϕ(|x|) con ϕ(x) = , x = 0. |x| |x|2 K K Una primitiva di ϕ(r) = 2 e` data da u(r) = − . Per la Proposizione 5.8 un r r K potenziale di F e` f (x) = − , x = 0. |x| Esempio 5.10 Sia F(x) = |x|8 x per ogni x ∈ R3 . Sicuramente F e` radiale su R3 \ {0} dato che x F(x) = ϕ(|x|) ∀x =  0, |x| con ϕ(r) = r9 . Un potenziale di F su R3 e` quindi f (x) = tratta in realt`a di un potenziale su tutto R3 .

1 10 |x| . Si noti che si 10

Campi conservativi e Test delle derivate (Sez. 5.3)  I campi conservativi continui sono tutti e soli i campi gradiente. Proposizione 5.11 Siano D ⊆ Rn aperto, e F : D → Rn un campo continuo. Allora F e` conservativo se e solo se F e` un campo gradiente. I campi conservativi di classe C 1 sono irrotazionali.

Note sul Capitolo 5

35

Definizione 5.12 (Campo irrotazionale) Sia F : D ⊂ Rn → Rn un campo vettoriale di classe C 1 , cio`e F = (F1 , ..., Fn ) con Fi : D → R di classe C 1 . Il campo F si dice irrotazionale se e solo se, per ogni i, j ∈ {1, ..., n},

∂xj Fi = ∂xi Fj . Esempio 5.13 In R3 il campo F(x) = (F1 , F2 , F3 )(x1 , x2 , x3 ) := (cos(x1 )x22 x3 , 2 sin(x1 )x2 x3 , sin(x1 )x22 ) e` irrotazionale. Il test prevede la verifica di tre uguaglianze:

∂x1 F2 (x) = 2 cos(x1 )x2 x3 = ∂x2 F1 (x), ∂x1 F3 (x) = cos(x1 )x22 = ∂x3 F1 (x), ∂x2 F3 (x) = 2 sin(x1 )x2 = ∂x3 F2 (x). Proposizione 5.14 (Campi conservativi e campi irrotazionali) Siano D ⊆ Rn aperto, e F : D → Rn un campo C 1 . 1. (Test sulle componenti) Se F e` conservativo allora F e` irrotazionale. 2. Se F e` irrotazionale e D e` semplicemente connesso allora F e` conservativo. Dimostrazione. Dimostriamo solo 1. Se il campo F e` conservativo esiste un potenziale f : D → R tale che ∇f = F. Per ogni i, j in 1, ..., n, per il Teorema di Schwarz si ha ∂xi Fj = ∂xi (∂xj f ) = ∂xj (∂xi f ) = ∂xj Fi .  Osservazione 5.15 Una palla e` un insieme stellato rispetto al centro, quindi un insieme semplicemente connesso: segue da 2. della Proposizione 5.14 che, localmente (almeno su una palla centrata in ogni punto del dominio), un campo irrotazionale e` ivi conservativo. Esempio 5.16 (Un campo irrotazionale che non e` conservativo) La conclusione della parte 2 della Proposizione 5.14 non vale in generale se D non e` semplicemente connesso. Si consideri ad esempio il campo definito su R2 \ {(0, 0)} da   x2 x1 , ∀(x1 , x2 ) = (0, 0). F(x1 , x2 ) = − 2 x1 + x22 x21 + x22 Si vede she F e` irrotazionale. Infatti     x1 x2 x22 − x21 ∂x2 − 2 = ∂ = x 1 2. x1 + x22 x21 + x22 (x21 + x22 )

36

Note sul Capitolo 5

Tuttavia F non e` irrotazionale. Infatti indicando con γ(t) = cos t, sin t), t ∈ [0, 2π] il circolo unitario positivamente orientato si ha   2π F · dr = 1 dt = 2π = 0. γ

0

Si noti che F, essendo irrotazionale, ammette primitive locali. Ad esempio si verifichi x1 che su x1 > 0 la funzione f (x1 , x2 ) = arctan e` un potenziale per F: la forma x2 differenziale x2 x1 − 2 dx1 + 2 dx2 2 x1 + x2 x1 + x22 associata al campo F e` quindi, localmente, il differenziale dell’angolo θ(x1 , x2 ) del punto (x1 , x2 ). Si pu`o intuire come non ci sia modo di estendere θ ad una funzione continua su tutto R2 \ {(0, 0)}: ad esempio percorrendo un giro sul circolo unitario π γ da (0, −1) con valore iniziale θ(0, −1) = − si torna al punto considerato con 2 π π lim θ(x1 , x2 ) = 3 = − . (x1 ,x2 )→(0,−1) 2 2 x21 +x22 =1,x1 0 si ottiene 1 f (x, y) = log(x2 + y 2 ) + g(y), 2 con g derivabile su R (attenzione: abbiamo scelto, ad esempio, y > 0 perch´e se y = 0 il campo F(x, 0) e` definito solo per x = 0: una funzione la cui derivata e` nulla su R \ {0} pu`o non essere costante!). La condizione

∂y f (x, y) =

x2

y + y2

porge allora che g e` costante su y > 0. Pertanto una primitiva di F su y > 0 e` 1 f (x, y) = log(x2 + y 2 ). Ora f e` definita su R2 \ {(0, 0)} e le regola della deri2 vazione mostrano che in realt`a ∇f = F su D: f e` quindi primitiva di F su tutto il dominio.

Note sul Capitolo 5

37

Formula di Green (Sez. 5.4) Se D e` una regione del piano per la quale vale la formula di Green, indicheremo con ∂ + D il suo bordo, orientato positivamente (percorso lasciando il braccio sinistro all’interno del dominio). Vediamo alcune applicazioni interessanti della formula do Green. Esempio 5.18 (Area di un dominio con frontiera in coordinate polari) Sia G un aperto regolare limitato, la cui frontiera sia parametrizzata da

γ(t) = ρ(t) (cos t, sin t), Proviamo che

1 Area (G) = 2



t2

t1

t ∈ [t1 , t2 ], ρ2 (t) dt.

La formula di Green porge Area (G) =

1 2

 ∂+G

x dy − y dx;

 x(t) = ρ(t) cos t, y(t) = ρ(t) sin t

e`

da cui Area (G) =

1 2



t2

t1

ρ(t) cos t(ρ (t) sin t + ρ(t) cos t)

− ρ(t) sin t(ρ (t) cos t − ρ(t) sin t) dt   1 t2 2 1 t2 2 = ρ (t)(cos2 t + sin2 t) dt = ρ (t) dt. 2 t1 2 t1 Esempio 5.19 (Area di un poligono) Chiamiamo poligono di R2 ogni sottoinsieme limitato di R2 la cui frontiera sia una poligonale semplice chiusa (una poligonale (a0 , ..., am ) si dice chiusa se a0 = am , semplice chiusa se gli unici lati non consecutivi che si intersecano sono il primo e l’ultimo, che si intersecano esattamente in a0 = am ). Detto ci`o, se ak = (ξk , ηk ) sono vertici successivi di una poligonale semplice chiusa, mostriamo che l’area del poligono racchiuso e` (si intende che sia ξm+1 = ξ1 , ηm+1 = η1 ) m   1   A =  (ξk−1 ηk − ξk ηk−1 )  2 k=1     m m   1  1    =  ηk (ξk+1 − ξk−1 ) =  ξk (ηk+1 − ηk−1 ) .   2 2 k=1

k=1

38

Note sul Capitolo 5

Tali formule di Gauss sono note in topografia. La frontiera del poligono G e` costituita dalla giustapposizione dei segmenti

γk (t) = ak + t(ak+1 − ak ) = (ξk + t(ξk+1 − ξk ), ηk + t(ηk+1 − ηk )), t ∈ [0, 1], k = 0, ..., m − 1. Per la formula di Green si ha      1 m−1  1    x dy − y dx =  x dy − y dx . Area (G) =   2 ∂+G 2 k=0 γk Il modulo qui appare perch´e non sappiamo se la giustapposizione dei cammini data costituisca una orientazione positiva del bordo di G. E`  γk

=

x dy − y dx 

1

[ξk + t(ξk+1 − ξk )](ηk+1 − ηk ) − [ηk + t(ηk+1 − ηk )](ξk+1 − ξk ) dt  1  1 = ξk (ηk+1 − ηk ) − ηk (ξk+1 − ξk ) dt = ξk ηk+1 − ξk+1 ηk dt 0

0

da cui

0

m−1  m     1 1      (ξk ηk+1 − ξk+1 ηk ) =  (ξk−1 ηk − ξk ηk−1 ) . Area (G) =    2  k=0 2 k=1

Ad esempio quanto vale l’area (0, −2), (3, 5), (−4, 8), (−1, −1)?

del

quadrilatero

di

vertici

Superfici e Aree (Sez. 5.5) Precisiamo meglio la definizione di superficie parametrica. Definizione 5.20 (Superficie parametrica) Una superficie parametrica in R3 e` una funzione vettoriale continua r : R → R3 ,

(u, v) → (f (u, v), g(u, v), h(u, v))

definita su una regione chiusa e limitata R del piano. L’insieme dei punti (x, y, z) = r(u, v), al variare di (u, v) ∈ R, si chiama sostegno della superficie parametrica; spesso, similmente a quanto fatto con le curve, nel Libro esso viene chiamato superficie di cui r e` una parametrizzazione.

40

Note sul Capitolo 5

da cui Area (Tf ) =

=

 R  b a



=

a

b

x2 (t) + y 2 (t)dt dz



f (x(t),y(t))

0



x2 (t) + y 2 (t) dz dt

 2 2 f (x(t), y(t)) x (t) + y (t) dt = f ds. γ

Vediamo cos’`e una superficie di rotazione ottenuta ruotando una curva attorno ad un asse coordinato. Proposizione 5.22 (Teorema di Pappo - Guldino per le superficie di rotazione) Sia γ =: [a, b] → [0, +∞[×{0} × R, γ(t) = (γ1 (t), 0, γ2 (t)) una curva di classe C 1 il cui sostegno e` contenuto nel semipiano xz con x ≥ 0. La funzione r(t, θ) = (γ1 (t) cos θ, γ1 (t) sin θ, γ2 (t)),

t ∈ [a, b], θ ∈ [0, 2π]

e` una superficie parametrica: essa viene detta la superficie di rotazione ottenuta ruotando (di 2π ) la curva γ attorno all’asse z . L’area di r vale  2π x ds = 2πxγ Lungh (γ), γ

dove xγ =

γ

x ds

Lungh (γ) Dimostrazione. Si ha

e` la distanza del baricentro di γ dall’asse di rotazione.

rt (t, θ) = (γ1 (t) cos θ, γ1 (t) sin θ, γ2 (t)) rθ (t, θ) = (−γ1 (t) sin θ, γ1 (t) cos θ, 0). Pertanto

|rt × rθ |2 (t, θ) = γ12 (t)γ12 (t) + 0 + γ12 (t)γ12 (t) = γ12 (t)|γ  (t)|2 , 

da cui Area =

[a,b]×[0,2π]



= 2π

a

b

γ1 (t)|γ  (t)| dt dθ

γ1 (t)|γ  (t)| dt = 2π



x ds. γ



Note sul Capitolo 5

41

Figura 5.3. Area di una superficie di rotazione

Osservazione 5.23 Si noti che i punti del sostegno di r si ottengono ruotando quelli del √ sostegno di γ attorno all’asse z . Infatti (x, y, z) sta nel sostegno di r se e solo se ( x2 + y 2 , 0, z) sta sul sostegno di γ . Esempio 5.24 (Toro vuoto) Un toro vuoto e` la superficie che si ottiene facendo ruotare (di 2π ) un circolo (bordo del disco) di raggio a > 0 attorno ad una retta che dista R > a dal centro del disco (una ciambella vuota, od una camera daria di pneumatico). Chiaramente il baricentro del circolo e` il centro del disco. Ne segue che l’area del toro vuoto e` 2πR(2πa) = 4π 2 aR. Formule simili si ottengono ruotando attorno ad un altro asse: ricordarsi, similmente

 al caso dei volumi, che interviene la distanza del baricentro della curva data dall’asse

di rotazione. Abbiamo visto sopra che il volume di un solido di rotazione si pu`o calcolare integrando le aree delle sezioni. Ci`o non accade in genere per le aree. Esempio 5.25 Facciamo ruotare il grafico di una funzione f : [a, b] → [0, +∞[ contenuto nel semipiano xz con z ≥ 0 attorno all’asse x. Il grafico di f e` il sostegno della curva γ(t) = (t, 0, f (t)); per il Teorema di Pappo - Guldino l’area della superficie ottenuta vale (si faccia attenzione che ora la distanza del baricentro di γ dall’asse di rotazione e` l’ordinata, e non l’ascissa, del baricentro di γ )  Area = 2π

γ



z ds = 2π

a

b

f (t) 1 + f  (t)2 dt.

42

Note sul Capitolo 5

Figura 5.4. Una superficie ottenuta ruotando il grafico di una funzione f attorno

all’asse x

Si osservi che le lunghezze delle sezioni sopra t hanno lunghezza 2πf (t), e si ha  Area =

a

b

2πf (t) dt

se e solo se f  ≡ 0 il che accade se e solo se la superficie e` un cilindro.

Esercizi Esercizio 5.1 Determinare l’area della superficie ottenuta ruotando la curva z = √ 2 x, 1 ≤ x ≤ 2 attorno all’asse x. Esercizio 5.2 Determinare l’area del cono ottenuto ruotando attorno all’asse z la curva z = 1 − x, 0 ≤ x ≤ 1. Esercizio 5.3 Determinare l’area della superficie ottenuta ruotando la curva z = x3 , 0 ≤ x ≤ 1/2 attorno all’asse x. Esercizio 5.4 Determinare l’area della superficie ottenuta ruotando la curva x = √ 15 attorno all’asse z . 2 4 − z, 0 ≤ z ≤ 4

Soluzioni degli esercizi Soluzione es. 5.1.

√ 8π √ (3 3 − 2 2). 3

...

(77,5+0*,

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Risposte agli esercizi dispari Capitolo 1 Esercizi di ripasso, pp. 28–30  

y

 

y

y y

y = 2x + 1 1

4y2 – 4x2 = 1 1

t=0 x2 + (y + 2)2 = 4

r = –4 sin  x

1 t=0 2 –1 2

x

0

0

1

x

x

–3 2

(0, –2)

y=– 3 2

y

y

  r = 2√2 cos 

y

r = – 5 sin 

y = x2 ⎛ ⎝√2

t=0

–1

1

0

⎛2 2 ⎛ x + ⎝y + 5 ⎝ = 25 2 4

⎛

, 0⎝

x

t=

1

⎛ ⎝x

x

⎛2

x

⎛ 0, ⎝

2

– √2⎝ + y = 2

⎛ – 5⎝ 2

y

y ⎛x ⎝

2 2 – 3⎛ + y = 9 2⎝ 4

0 ≤ r ≤ 6 cos 

r = 3 cos  ⎛3 ⎛ ,0 ⎝2 ⎝

y

x 6

0

x

y x=2 x – √3 y = 4√3

H y

y x 4√3

2

x 3

–4 2 0

r=

(1, 0)

2 1 + cos 

r=

(2, ) x (6, )

0

–2 –3

6 1 – 2 cos  x

9  9PZWVZ[LHNSPLZLYJPaPKPZWHYP

Capitolo 2

Esercizi di ripasso, pp. 61–64

y 2 √2 1

v ⎞⎠ ⎞⎠ 4 a ⎞⎠ ⎞⎠ 4

0

–4

(2√2, 1) v(0)

a(0)

4

x

–1

,Q 



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Capitolo 3

Esercizi di ripasso, pp. 130–134

y

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z=1 x

y 3 z=9 –1

1

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1 0

1

s

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0

3

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