Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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-0 PROBLEMS PLUS 1. Let 81 be the portion ofD(8) between 8(a) and 8, and let 881 be its boundary. Also let 8L be the lateral surface of 81 [that is, the surface of 8 1 except 8 and 8(a)]. Applying the Divergence Theorem we have/" { But r ·3n d8 = { { { \l. r 3 dV. J 8S1 r J J J s 1 r "il · :3 = \ :x 1 :y'!) · \ (x2 + y2x+ z2)3/2' (x2, + y2y + z2)3/2' (x2 + y./+ z2//2) (x2 + 1/ + z2 - 3x2) + (x2 + y2 + z2- 3y2) + (~2 + y2 + z2 - 3z2) - - o - (x2 + y2 + z2)5/2 - => j" { r ' 3 ° d8 = !"{ 0 dV = 0. On 'the other hand, notice that for the surfaces of 881 other than S(a) and 8, l os , r J l s, r · n = O => { r ·3n d8 => l os, r Js r Js(a) r JsL r i s r Js(a) r 0 = !"{ r ·3n dS = !"{ r ·3n dS + !"{ r ·an d8 + !"{ r ·sn d8 = !"{ r ·3n d8 + !" !Is !"1 - r · . - n r · n r r 3 d8 = - - 3 - d8. Notice that on S(a), r =a => n = - - =-- and r · r = r 2 = a 2 , so s 1 S(a) r r a . that - / " r r ·3n d8 = !" r r ·4r d8 = !" r a: d8 = ~ !" r dS = area of2S (a) = ID(8) 1. J s(a) r . Js(a) a J s(a) a a Js(a) a ! "{ Therefore ID(S)I = Js ~ r · n dS. 3. The given line i ntegra l ~ .{ 0 (bz - CIJ) dx +(ex- az ) dy + (ay - bx) dz can be expressed as fc F · dr if we define the vector field F by F (x, y, z) =Pi + Q j + R k = ~(bz- CIJ) i + Hex- az) j + ~(ay- bx) k. Then define 8 to be the planar interior ofC, so Sis an oriented, smooth surface. Stokes' Theorem says .fc F · dr = .ffs curlF · dS = JJ~ curl F . n d8. Now curl F = ( ~: - ~~) i + (a:; - ~=) j + ( ~~ - ~:) k = (~a+ ~a) i + (~b+ tb) j + (~c+ tc) k = a i +bj + ck = n so curlF · n = n · n = lnl 2 = 1, hence Jfs curl F · n dS = Jfs dS w hich is simply the su rface area of S. Thus, J~ F · dr = ~ j~(bz- cy) dx +(ex- az) dy + (ay .- bx) dz is the plane area.enclosed by C. 5. (F ·\l) G= (p~:.rc+Ql:y+R l !)cP2 i +Q2j+R2 k) = (p 1 8P2 + Q 1 8P2 + R 1 8P2) i + (PI 8Q2 + Q 1 8Q2 + R 1 8Q2) j ax By 8z OX 8y az = (F · \JP2) i + (F · \JQ2) j + (F · \JR2) k. ® 2012 Ccogagc Lc3ming. All Rights Rc:scr\'00. Moy not be SC30Ded, copied, or duplicated. or posted to a publicly accessible wcbsilc, in whole or in pan. 343
344 0 CHAPTER 16 PROBLEMS PLUS Similarly, (G · V) F = (G · V Pt) i + (G · VQt)j + (G · V Rl) k. Then F xcurlG = 8R2j8y - 8Q2/8z 8P2/Bz - 8R2/8x 8Q2j8x- 8P2/8y = (Q 1 8Q2 _ Q 1 8P2 _ Rt 8P2 + R t BR2) i + (R 1 BR2 _ Rt 8Q2 _ Pt 8Q2 + Pt BP2) j ax By Bz Bx By Bz Bx By k + (p 1 8P2 _ Pt 8R2 _ Q 1 BR2 + Q 1 BQ2 ) k 8z ox By Bz and Then and Hence (F · V) G + F x curl G + ( G · V) F + G x curl F = [(p 1 8P2 + p 2 BPt ) + (Q 1 BQ2 +Q 2 BQt) + (Rt BR2 + R2 BRt)] i ~ ~ ~ ~ ~ ~ @ 2012 Cengage Learning. All Rights Rcscn•ed. M::~y not be scanned, copied 1 or duplicated, or p
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.1 CURVES DEFINED BY PARA
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SECTION 10.2 CALCULUS WITH PARAMETR
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SECTION 10.3 POLAR COORDINATES 0 13
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SECTION 10 .~ POLAR COORDINATES 0 1
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SECTION 10.4 A~~S AND LENGTHS IN PO
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SECTION 10.4 AREAS AND LENGTHS IN P
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SECTION 10.4 AREAS AND LENGTHS IN P
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SECTION 10.5 CONIC SECTIONS 0 27 5.
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SECTION 10.5 CONIC SECTIONS 0 29 35
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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SECTION 10.6 CONIC SECTIONS IN POLA
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CHAPTER 10 REVIEW 0 35 the length o
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CHAPTER 10 REVIEW 0 37 EXERCISES 1.
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CHAPTER 10 REVIEW 0 39 25. x = t +
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CHAPTER 10 REVIEW 0 41 2 2 . 45. ~
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0 PROBLEMS PLUS l lt sin u dx cost
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11 . D INFINITE SEQUENCES AND SERIE
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SECTION 11.1 SEQUENCES 0 47 35. a,.
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SECTION 11.1 SEQUENCES D 49 71. Sin
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SECTION 11.2 SERIES 0 51 ak+l- a1.:
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. . (1+ c)- 2 scn es and set 1t equ
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SECTION 11.3 THE INTEGRAL TEST AND
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, SECTION 11.3 THE INTEGRAL TEST AN
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SECTION 11.4 THE COMPARISON TESTS D
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SECTION 11.5 ALTERNATING SERIES 0 6
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SECTION 11.5 ALTERNATING SERIES D 6
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SECTION 11.6 ABSOLUTE CONVERGENCE A
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17 lim I an+l I= SECTION 11.7 STRAT
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SECTION 11.9 REPRESENTATIONS OF FUN
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SECTION 11.10 TAYLOR AND MACLAURIN
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61 _ x_ x · sin x - x- tx 3 + 1~
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SECTION 11.11 APPLICATIONS OF TAYLO
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CHAPTER 11 REVIEW 0 97 J'(xn)(xn -
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CHAPTER 11 REVIEW 0 99 oo f (n) (O)
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CHAPTER 11 REVIEW 0 101 l 23. Consi
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49./- 1 - dx = -ln{4- x) + C and 4-
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D PROBLEMS PLUS 1. It would be far
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CHAPTER 11 PROBLEMS PLUS D 107 l 9.
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112 0 CHAPTER 12 VECTORS AND THE GE
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D PROBLEMS PLUS 1. Since three-dime
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l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
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13 D VECTOR FUNCTIONS 13.1 Vector F
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.1 VECTOR FUNCTIONS AND S
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SECTION 13.2 DERIVATIVES AND INTEGR
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2 SECTION 13.4 MOTION IN SPACE: VEL
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CHAPTER 13 REVIEW D 173 43. The tan
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5. f~(t 2 i +tcos 1rtj +sin 1rt k )
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CHAPTER 13 REVIEW 0 177 23. (a) r (
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180 0 CHAPTER 13 PROBLEMS PLUS (-2a
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184 0 CHAPTER 14 PARTIAL DERIVATIVE
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D PROBLEMS PLUS 1. The areas of the
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15 D MULTIPLE INTEGRALS 15.1 Double
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SECTION 15.2 ITERATED INTEGRALS D 2
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l SECTION 15.3 DOUBLE INTEGRALS OVE
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SECTION 15.3 DOUBLE INTEGRALS OVER
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61 . Since m :S j(x, y) :S M, ffD m
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SECTION 15.5 APPLICATIONS OF DOUBLE
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SECTION 15.5 APPLICATIONS OF DOUBLE
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SECTION 1505 APPLICATIONS OF DOUBLE
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-I SECTION 15.6 SURFACE AREA 0 267
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Therefore E = { (x, y, z) I -2 ~ X~
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43. I,. = foL foL foL k(y2 + z2)dz.
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SECTION 15.8 TRIPLE INTEGRALS IN CY
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M xv = I~1f I: I:2 6 - 3 r 2 (zK) r
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SECTION 15.9 TRIPLE INTEGRALS IN SP
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(b) The wedge in question is the sh
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SECTION 15.10 CHANGE OF VARIABLES I
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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