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

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. . . y2 2 33. An equivalent equation IS - x 2 + 4 -;- z = 1, a 35. Completing the squate in y gives hyperboloid of two sheets with axis the y-ax.is. For IYI > 2, traces parallel to the xz-plane are circles. z CHAPTER 12 REVIEW 0 145 z2 4x 2 + 4(y - 1) 2 + z 2 = 4 or x 2 + (y - 1) 2 +- = 1, 4 . an ellipsoid centered at (0, 1, 0). (0, 1,2) (O, l ,-2) . x2 y2 2 2 2 4 + 16 = l. The equation of the ellipsoid is ~ + i 6 + : 2 = l, since the horizontal trace in the plane z = 0 must be the original ellipse. The traces of the ellipsoid in the yz -plane must be circles since the surface is obtained 2 2 2 by rotation about the x-axis. Therefore, c 2 = 16 and the equation of the ellipsoid is ~ + i 6 + ~ 6 = 1 4x 2 + y 2 + z 2 = 16. @ 2012 Cengngc lc;1ming. All Rights ReserVed. Muy nol be sc3nncd, corictl. or duplicated, or posted to u publicly ncccssibh.: \\'cbsite, in whole or in par1.
D PROBLEMS PLUS 1. Since three-dimensional situations are often difficult to visualize and work with, let us first try to find an analogous problem in two dimensions. The analogue of a cube is a square and the analogue of a sphere is a circle. Thus a similar problem in two dimensions is the following: if five circles with the same radius rare contained in a square of side 1 m so that the circles touch each other and four of the circles touch two sides ofthe square, find r. The diagonal of the square is .J2. The diagonal is a lso 4r + 2x . But x is the diagonal of a smaller square of side r . Therefore x = .J2r ~ .J2=4r+2x =4r+2.J2r= (4+2.J2)r ~ r = 4 /{,/2. Let's use these ideas to solve the original three-dimensional problem. The diagonal of the cube is \/12 + 12 + 12 = v'a. The diagonal of the cube is also 4r + 2x where x is the diagonal of a smaller cube with edger. Therefore r;; r;; v'3 2 v'a - 3 x = Jr 2 + 1·~ + r 2 = v'3 r ~ v'a = 4r+2x .=4r+2v3r = (4+2v3)r.Thus r= r;;= 4 + 2v3 2 The radius of each ball is ( v'3 - ~) m. 3. (a) We find the line of intersection L as in Example 12.5. 7(b). Observe that the point ( - 1, c, c) lies on both planes. Now since L lies in both planes, it is perpendicular to both of the normal vectors n1 and n 2, and thus parallel to their cross product j k n 1 x n 2 = c 1 1 = (2c, -c 2 + 1, -c 2 - 1). So symmetric equations of L can be written as 1 - c c -- x +1 = y - c z - c ~ 1 = - 2c (..- - ~ c + 1 , provided that c I 0, ±1. If c = 0, then the two planes are given by y + z = 0 and x = - 1, so symmetric· equations of L are x = - 1, y = - z. If c = - 1, then the two planes are given by -x + y + z = - 1 and x + y + z = -1, and they intersect in the line x = 0, y = - z - 1. If c = 1, then the two planes are given by x + y + z = 1 and x - y + z = 1, and they intersect in the line y = 0, X = 1- Z. (b) If we set z = t in the symmet~i c equations and solve for x andy separately, we get x + 1 = (t - c)( - 2 c) c 2 + 1 (t- c)(c 2 - 1) y - c = c2 +1 - 2ct + (c 2 - 1) (c 2 - 1)t + 2c El. . . fr th . ~ x = c 2 + 1 , Y = & + 1 . unmatmg c om ese equattons, we have x 2 + y 2 = t 2 + 1. So the curve traced out by L in the plane z = tis a circle with center~~ (0, 0, t) and radius J fl + 1. © 2012 Cc:ngoge le=ling. All Rights Resorvoo. Moy nol be sconned, copied, or duplicalcd, or posll:d to o publicly occcssiblc wcbsito, in whole or in pori. 147
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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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SECTION 11.2 SERIES 0 53 13. n s.,
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SECTION 11.2 SERIES 0 55 47. F~r th
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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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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 11.8 POWER SERIES 0 75 l 00
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SECTION 11.8 POWER SERIES 0 n (b) I
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SECTION 11.9 REPRESENTATIONS OF FUN
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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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Page 155 and 156: l CHAPTER 12 PROBLEMS PLUS 0 149 Eq
Page 157 and 158: 13 D VECTOR FUNCTIONS 13.1 Vector F
Page 159 and 160: SECTION 13.1 VECTOR FUNCTIONS AND S
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198 D CHAPTER 14 PARTIAL DERIVATIVE
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200 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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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 15.4 DOUBLE INTEGRALS IN PO
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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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SECTION 15.7 TRIPLE INTEGRALS 0 269
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SECTION 15.7 TRIPLE INTEGRALS D 271
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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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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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CHAPTER 15 REVIEW 0 289 15 Review C
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CHAPTER 15 REVIEW 0 291 l 9. The vo
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CHAPTER 15 REVIEW D 293 33. Using t
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49. Since u = x- y and v = x + y, x
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298 D CHAPTER 15 PROBLEMS PLUS To e
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300 0 CHAPTER 15 PROBLEMS PLUS 13.
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16 0 VECTOR CALCULUS 16.1 Vector Fi
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SECTION 16.2 LINE INTEGRALS 0 305 2
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SECTION 16.2 LINE INTEGRALS 0 307 (
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SECTION 16.2 LINE INTEGRALS D 309 3
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SECTION 16.3 THE FUNDAMENTAL THEORE
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SECTION 16.4 GREEN'S THEOREM D 313
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SECTION 16.4 GREEN'S THEOREM 0 315
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8 8 8 (b)clivF = 'V ·iF=- (x+yz) +
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SECTION 16.5 CURL AND DIVERGENCE 0
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SECTION 16.6 PARAMETRIC SURFACES AN
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that is, D = {( x, y) I x 2 + y 2 :
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SECTION 16.7 SURFACE INTEGRALS 0 32
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SECTION 16.7 SURFACE INTEGRALS 0 33
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SECTION 16.8 STOKES' THEOREM 0 333
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dS SECTION 16.9 THE DIVERGENCE THEO
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CHAPTER 16 REVIEW 0 337 27. JI 5 cu
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CHAPTER 16 REVIEW 0 339 TRUE-FALSE
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CHAPTER 16 REVIEW D 341 Alternate s
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344 0 CHAPTER 16 PROBLEMS PLUS Simi
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3.c6 0 CHAPTER 17 SECOND-ORDER DIFF
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348 0 CHAPTER 17 SECOND-ORDER DIFFE
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352 D CHAPTER 17 SECOND-ORDER DIFFE
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356 D CHAPTER 17 SECOND-ORDER DIFFE
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0 APPENDIX Appendix H Complex Numbe
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APPENDIX H COMPLEX NUMBERS 0 361 43
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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