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# 15 - CHAPTER 15 Vector Analysis Section 15.1 Vector Fields...

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Unformatted text preview: CHAPTER 15 Vector Analysis Section 15.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 Section 15.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 Section 15.3 Conservative Vector Fields and Independence of Path . . . . . . 387 Section 15.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 395 Section 15.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 404 Section 15.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 412 Section 15.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 423 Section 15.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 428 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 CHAPTER Vector Analysis Section 15.1 15 Vector Fields 2. All vectors are parallel to x-axis. Matches (d) 5. Vectors are parallel to x-axis for y n. Matches (a) 3. All vectors point outward. Matches (b) 6. Vectors along x-axis have no x-component. Matches (f) 9. F x, y F y 1. All vectors are parallel to y-axis. Matches (c) 4. Vectors are in rotational pattern. Matches (e) 7. F x, y F i 2 y j 8. F x, y F 2 2i xi x2 yj y2 c x2 y2 5 c2 y 1 1 −4 x −2 −1 1 2 x −5 −4 −2 x 3 5 −5 10. F x, y F xi x2 y 5 4 3 2 1 yj y2 11. F x, y, z F 3y z 4 3y j c 12. F x, y F x xi c y 3 2 1 −5 − 4 −2 −1 −2 −3 −4 −5 x 12 45 x 2 4 y −3 −2 −1 1 2 3 x 13. F x, y F x2 c 2 16 4x i 16x2 y2 c2 y 2 yj y2 1 c 14. F x, y F y 10 8 6 4 x2 1 y2 i x2 y2 j 2 15. F x, y, z F 3 z i j k 4 4 −4 y −2 −1 x 1 2 −2 x 2 4 −4 x −2 363 364 Chapter 15 xi x2 y2 z2 z Vector Analysis yj zk z2 c 17. 2 1 x 1 −1 2 16. F x, y, z F x2 y 18. F x, y 6 4 2y y 3x i 2y 3x j y2 c2 −2 −1 2 −6 −4 −2 x 2 4 6 2 −2 −2 −2 −6 2 x −2 2 y 19. 2 1 z 20. F x, y, z z xi yj zk 21. f x, y fx x, y fy x, y F x, y 5x2 10x 3x 10x 3xy 3y 20y 3y i 10y 2 2 1 1 2 x 1 2 y 3x 20y j 1 2 x 1 2 y 22. f x, y fx x, y fy x, y F x, y sin 3x cos 4y 3 cos 3x cos 4y 4 sin 3x sin 4y 3 cos 3x cos 4y i 4 sin 3x sin 4yj 23. f x, y, z fx x, y, z fy x, y, z fz F x, y, z z ye x 2 2 x ye x ex 1 2 2 2 x ye x i 2 ex j 2 k 24. f x, y, z fx x, y, z fy x, y, z fz x, y, z F x, y, z y z z x2 1 z z x z y xz y2 y z2 z x2 xz y 1x x y z i y 1 z xz j y2 y z2 1 x x k y 25. g x, y, z gx x, y, z gy x, y, z gz x, y, z G x, y, z x y ln x y ln x x ln x 0 xy x y y y y xy x xy x y y y ln x yi xy x y x ln x yj S ection 15.1 26. g x, y, z gx x, y, z gy x, y, z gz x, y, z G x, y, z x arcsin yz arcsin yz xz 1 xy 1 y 2z2 xz 1 y i x2 yz 1 j x 22 Vector Fields 365 27. F x, y 12xy i 6 x2 6 x2 yj y have continuous first M 12 xy and N partial derivatives. N x 12x y 2z2 M ⇒ F is conservative. y arcsin yz i j xy 1 y 2z2 k 28. F x, y 1 yi x2 xj 29. F x, y sin y i x cos y j x cos y have continuous first partial M y x2 and N derivatives for all x N x 1 x2 1 x have continuous first partial 0. M sin y and N derivatives. N x cos y M ⇒ F is conservative. y 1 yi xy 1 i x 1 j y M ⇒ F is conservative. y 30. F x, y xj 31. M N x 15y3, N 5y 2 5xy 2 M y 45y 2 ⇒ Not conservative M 1 x and N 1 y have continuous first partial derivatives for all x, y 0. N x 0 M ⇒ F is conservative. y x x2 x2 y2 y x2 32 32. M N x ,N xy y2 y2 M ⇒ Conservative y 33. M N x 2 2x y e ,N y 2x 2 x e y2 y y 2 y 2x 2x e y3 M ⇒ Conservative y 34. M N x y 1 1 x2y2 ,N x 1 x2y2 32 35. F x, y 1 x2y2 y x 2 xy x2 2 xy i 2x 2x x2 j 1 x2y2 M y 1 32 ⇒ Not conservative Conservative fx x, y 1 yi y2 1 i y 1 yy x 2x y2 1 y2 2 y2 2 xy, fy x, y x 2, f x, y x2y K 36. F x, y 2xj 2x j y2 37. F x, y y x 2 xye x x 2e x 2 xe x y 2y i 2 2 xj 2 38. F x, y 2 x 3 ye x 2 3 x2 y 2 i 6x2 y 6x2y 2x3yj y 2 xe x 2 xe x 2 y y y x 3 x2 y2 2x3y y y 2 x 3 ye x 2 y Conservative fx x, y fy x, y f x, y 2 xye x 2e x ex 2 2 Conservative x2y y fx x, y fy x, y f x, y 3 x2 y 2 2x3y x3y 2 K Not conservative y K 366 Chapter 15 x x2 x y x2 y x x2 y2 y2 y2 Vector Analysis y x2 x2 x2 2 xy y2 2 xy y2 y2 2 39. F x, y i j 40. F x, y 2y yx x x2 y2 2y i x 2 x x2 j y2 41. F x, y y e x cos y i sin y j e x cos y e x sin y e x sin y e x sin y 2 2x y2 x Conservative fx x, y fy x, y f x, y x x2 y x2 y2 y2 K 1 ln x2 2 2x x2 2x y x2 x x2 y2 2y y2 2 2 Not conservative y2 Not conservative 42. F x, y y2 2 i 2y x2 8 xy x2 y 2 8 xy x2 y2 3 y2 2 j 43. F x, y, z x yz i i j yj k z z k z k, 1, 2, 1 curl F 3 xy xyz y 2j xyj xz k curl F 1, 2, 1 Conservative fx x, y fy x, y f x, y 2x x2 2y x2 1 x2 x2z i i curl F x 2z x z z curl F 2, 46. F x, y, z j y2 y2 2 y2 2 K 44. F x, y, z 2 xz j k y z k, 2, 1, 3 45. F x, y, z e x sin y i i e x cos y j, 0, 0, 3 j y x cos y e 2k k z 0 2e x cos y k y z yz 2xz 2x i 2x i 0 x2 j 7i i j y e 6i x yz curl F x x sin y e x2 j 2z k 4j 6k 2z 0k curl F 0, 0, 3 1, 3 e i x yz j k , 3, 2, 0 k z e 6j x yz curl F e x x yz xz xy e x yz i yz xy e x yzj yz xz e x yz k curl F 3, 2, 0 S ection 15.1 x i y Vector Fields 367 47. F x, y, z arctan i ln x 2 j y y2 j k z y2 1 k curl F x x arctan y yz y i z x x2 y2 1 x y2 xy 2 k 2x x2 y2 k 1 ln x 2 2 xz x j y xz z 48. F x, y, z i j k z xy xy x y k curl F y x yz z x2 x x2 x sin x i y 1 y 2 x zx x2 x 2 y z 2 i i y2 xy y2 x y2 2 y y 2 z y 2 j j z2 xz 2 2 z2 yz 1 z 2 2 k 1 k 1 x z 2 1 1 z z2 y x z 2 49. F x, y, z yi sin y j y zj k z sin z xk curl F sin x 50. F x, y, z x y sin y y2 i cos y x zi cos z xj cos x yk z sin z j j y y2 k k x2 z2 i k y zi z y2 z x2 xj x y 2 z2 yk curl F x2 51. F x, y, z sin y i i curl F x sin y x y2 z2 x cos y j j k x2 z2 x2 z2 52. F x, y, z ez y i i j xj k xe z i yezj 0 k y z x cos y 1 2 cos yk 0 curl F x yz yez xez ez Not conservative 53. F x, y, z ez y i i curl F j xj k 0 xyk Not conservative 54. F x, y, z y 2z 3 i i curl F y x 2z 3 2 xyz 3 j j k 3 xy2z 2 k x y z yez xe z xye z y z 2 xyz 3 3xy 2z2 0 Conservative fx x, y, z fy x, y, z fz x, y, z f x, y, z ye z xe z x ye z x ye z K Conservative f x, y, z x y 2z3 K 368 Chapter 15 1 i y i curl F x 1 y j y x y2 Vector Analysis x j y2 k z 2z 1 0 curl F x2 Conservative 1 y x y2 2z 1 x y x y g y, z h x, z K1 K2 f x, y, z K3 fx x, y, z fy x, y, z fz x, y, z f x, y, z x x2 y x2 1 x x2 1 ln x 2 2 y x2 1 ln x 2 2 f x, y, z f x, y, z dz 1 ln x 2 2 xex i x xe x ex x xex ex 1 yey j y ye y ey y yey ey 1 z y2 y2 dx y2 dy y2 h x, z K3 K K2 g y, z K1 y2 y2 x x2 i x x y2 x2 y2 y x2 j y y y2 y2 k z 1 0 55. F x, y, z 2z 1k 56. F x, y, z i j k Conservative fx x, y, z fy x, y, z fz x, y, z f x, y, z f x, y, z f x, y, z z2 f x, y, z x y 1 dx y x dy y2 2z z z2 1 dz p x, y z K p x, y y2 z 57. F x, y, z div F x, y, z 6x 2 i x 6x 2 xy2j y 2 xy x y2 58. F x, y div F x, y 12 x 59. F x, y, z div F x, y, z sin x i x sin x cos y j y y2 i y2 z2 k cos y z ln y 2 y xy z2 cos x sin y 2z 60. F x, y, z div F x, y, z ln x 2 x xy j z2 k z ln y 2 z2 2x x2 62. y2 x 2z y2 z2 x 2z i 2 xz 11 2 xz j y yz k ln x 2 61. F x, y, z div F x, y, z div F 1, 2, 1 x yz i yz 4 1 yj 1 zk yz 2 F x, y, z div F x, y, z div F 2, 1, 3 S ection 15.1 63. F x, y, z div F x, y, z div F 0, 0, 3 e x sin y i e x sin y 0 e x cos y j e x sin y 64. F x, y, z div F x, y, z div F 3, 2, 1 ln x yz i j k 111 x y z 11 11 1 32 6 Vector Fields 369 65. See the definition, page 1054. Examples include velocity fields, gravitational fields and magnetic fields. 66. See the definition of Conservative Vector Field on page 1057. To test for a conservative vector field, see Theorem 15.1 and 15.2. 68. See the definition on page 1062. 67. See the definition on page 1060. 69. F x, y, z G x, y, z F G i 1 x i xi 2x j yj 3y k zk 2 xz i curl F G 2 xz 70. F x, y, z G x, y, z F G xi x2 i ij x0 x2 y i curl F G x yz x xx 71. F x, y, z x yz i i curl F x xyz j y y i curl curl F x 0 zk yj k z z2 z2 k yz i j y xz2 2 xz 2z yj k z z j y xy xyj xz k x 2z x2 i 1i zk xz2 k z xy y zz yj 2x 1k z2 2 xz x2z j xyk x 3y 2 3xy 3y 2 i j y z y z k z 2x 2 1 3x y j y j k 2x 3y yz 2x2 k 1i 4x 2x j 3y 6y k 6xj 3y k zk 72. F x, y, z x2z i i j 2 xz j k z yz j yz k curl F x x2z y 2xz i z 2x i x2 j 2z k k z xz zj yk curl curl F z k z 2z j 2 xk x 2x y x2 370 Chapter 15 i xi Vector Analysis 2x j yj 3y k zk 74. F x, y, z G x, y, z F z 3x y j y 2x2 k G G xi xi i x x2 j 0 y 0 2 73. F x, y, z G x, y, z F G zk yj k z z2 z2k yzi xz2 x 2z j xyk ij k 1 2x 3y x yz 2 xz 3y 2 i 2z x yz i i j y y x Nj M x 3x yj k z z 0 P k and G Qi i N Qi Rj j y Q P N S N i z curl G R z N P x P Rj P k z S Ri M j z x P S M k y xyj xzk zk div F div F 75. F x, y, z G 76. F x, y, z x2z i i j 2 xz j k yz k curl F x xyz curl F x x2z 2 y z 2xz yz 2 0 z 2x i x2 j 2z k div curl F 77. Let F F Mi G div curl F S k where M, N, P, Q, R, and S have continuous partial derivatives. Sk curl F G M y P y x z M S y Qj R i z x N S x R Q j z y M R x Qk Q k y N x curl F 78. Let f x, y, z be a scalar function whose second partial derivatives are continuous. f f i x f j y i curl f x f x 79. Let F Mi G Nj j y f y f k z k z f z Ri y N Sj S T k. z P T M x M x div F R x N y div G N y P z S y R x P z S y T z T z 2f yz 2f i zy 2f xz 2f j zx 2f xy 2f k yx 0 P k and G x M R div F S ection 15.1 80. Let F F Mi G Nj P k and G NT Ri PS i Sj T k. MT PR j MS NR k Vector Fields 371 ijk MNP RST G x N NT T x P y curl F div F PS T N x N R z G y P PR S x M z S MT P x P S x P z R y MS R N x NR P y M T y M T M y T y M S z S z N S M z R z N T x R z P R N z S x R y M T y F curl G 81. F Mi f Nj Pk F curl curl curl f f F F curl F F (Exercise 77) (Exercise 78) 82. Let F Mi Nj i Pk . j k fF xyz fM fN fP f P y P y f P y N i z f N z P x f N i z M j z f P x N x f P x M k y f M z i f x f M j z jk ff yz P f N x f N x f M y f M k y f f F f F MN 83. Let F Mi Nj x P k, then f F fM y fN fMi z f Nj fP f f f Pk. M x M x M f x N y f F f N y N z N f y f M x div f F f P z f N y P f z f P z f div F 84. Let F Mi curl F div curl F Nj Pk. P y x 2P xy N i z P y N z 2N xz P x y 2P yx M j z P x 2M yz N x z 2N zx M k y N x 2M zy M z M y 0 (since the mixed partials are equal) 372 Chapter 15 Vector Analysis xi yj z k and f x, y, z F x, y, z x2 y2 z 2. In Exercises 85-88, F x, y, z 1 ln x 2 2 x x2 y2 1 y2 x x2 y2 z2 n 85. ln f ln f y2 z2 i z2 x2 y y2 z2 j x2 z y2 z2 k xi x2 yj y2 zk z2 F f2 86. 1 f 1 f x2 z2 3 2i y x2 y2 z2 3 2j z x2 y2 z2 3 2k xi x2 yj y2 zk z2 3 F f3 87. f n fn n x2 x2 y2 y2 n z2 z2 x2 z2 1 y2 n 1 x2 y2 n 2 x y2 n 1 z2 x2 i z y2 n x2 k y2 z2 n 1 x2 y y2 z2 j z2 xi z2 n 1 f x2 y2 yj zk nf n 2F 88. w w x w y w z 2w x2 x x2 x2 x2 y2 y y2 z y2 z2 x2 2w 2x 2 y 2 z 2 x2 y2 z2 5 2 2y 2 x 2 z 2 x2 y2 z2 5 2 2z 2 x 2 y 2 x2 y2 z2 5 2 2w 2w 2w z2 3 2 z2 3 2 z2 3 2 y2 2w z2 2w x2 y2 z2 0 Therefore w 89. True. F x y 16x2 y 4 → 0 as x, y → 0, 0 . 0 and y > 0. Thus, 1 is harmonic. f 90. True. If x, y is on the positive y-axis, then x F x, y F 0, y y 2j. 91. False. Curl is defined on vector fields, not scalar fields. 92. False. See Example 7. Section 15.2 Line Integrals 373 Section 15.2 1. x2 x2 9 cos2 t y2 y2 9 sin2 t cos2 t sin2 t x y rt 9 1 1 x2 9 y2 9 Line Integrals 2. cos2 t x2 16 y2 9 sin2 t cos2 t sin2 t x y rt 3 sin t j 1 1 x2 16 y2 9 4 cos t 3 sin t 4 cos t i 0≤t≤2 3 sin t j 3 cos t 3 sin t 3 cos t i 0≤t≤2 t i, 3i t 3 j, 9 t i 3 j, 12 t j, 0 3 6 9 ≤ ≤ ≤ ≤ t t t t ≤ ≤ ≤ ≤ 3 6 9 12 3. r t 4. r t ti 5i 14 4 5 t j, 0≤t≤5 9 t j, 5 ≤ t ≤ 9 t i, 9 ≤ t ≤ 14 5. r t ti 2 4ti x y ds t j, ti 2 0≤t≤1 t j, 1 ≤ t ≤ 2 4i 3 2 6. r t ti 4 8 t 2 j, ti t j, 0≤t≤2 4 j, 2 ≤ t ≤ 4 4≤t≤8 7. r t 3t j, 0 ≤ t ≤ 2; r t 2 3j 2 4t 0 3t 4 2 dt 0 5t dt C 5t 2 2 2 10 0 8. r t ti 4 xy d s 2 2 t j, 0 ≤ t ≤ 2; r t 4t 2 t 1 1 dt i 42 j 2 2t 0 t 2 dt 4 2 t2 C 0 t3 3 2 4 24 0 8 3 16 2 3 9. r t x2 C sin t i y2 cos t j z2 ds 8 t k, 0 ≤ t ≤ 2 2 ;rt cos t i 64 t 2 sin t j 2 8k sin t 2 sin2 t 0 2 cos2 t 64t 2 d t cos t 64 dt 8 2 3 3 65 1 0 65 t 64 t 3 3 2 65 0 65 6 3 16 2 10. r t 12 t i 8 xyz d s 5t j 2 3 k, 0 ≤ t ≤ 2; r t 122 52 12 i 02 d t 5j 2 8 12t 5t 3 0 18,720 t 2 d t 0 18,720 C t3 3 2 49,920 0 374 11. r t Chapter 15 Vector Analysis y t i, 0 ≤ t ≤ 3 3 x2 C y 2 ds 0 3 t2 t 2 dt 0 02 2 1 0 dt 1 x 1 2 3 13 t 3 12. r t x2 C 3 −1 9 0 t j, 1 ≤ t ≤ 10 10 10 y y2 ds 1 10 0 t2 1 t2 dt 10 0 1 dt 8 6 4 2 13 t 3 333 1 −4 −2 x 2 4 13. r t x2 C cos t i y2 ds sin t j, 0 ≤ t ≤ 2 y 2 1 cos 2 t 0 2 sin2 t sin t 2 cos t 2 d t x dt 0 2 1 14. r t x2 C 2 cos t i y2 ds 2 sin t j, 0 ≤ t ≤ 2 y 2 2 4 0 2 cos2 t 4 4 sin2 t 2 sin t 2 2 cos t d t 1 2 8 dt 0 x 1 2 15. r t x C ti t j, 0 ≤ t ≤ 1 1 y 4 y ds 0 t t2 2 4t 83 t 3 1 1 2 0 1 dt 19 2 6 1 (1, 1) 2 x 1 16. r t x C ti 3 t j, 0 ≤ t ≤ 3 3 9 y (3, 9) 4 y ds 0 t 4 3t t2 2 1 9 dt 6 10 8 33 t 3 144 3 2 0 −3 3 x 3 6 10 27 6 57 10 2 S ection 15.2 t i, 2 3 x C1 Line Integrals 375 17. r t ti t j, t 1 0≤t≤1 1 j, 1 ≤ t ≤ 2 2≤t≤3 t dt 0 2 y (0, 1) C2 4 y ds 4 y ds 1 1 2 t t2 2 t dt 8 3 2 4 6 8 4t 8 t 3 1 1 1 2 32 1 C3 x C2 2 2 2t 3 1 dt 19 2 6 3 32 2 C1 (1, 0) x x C3 4 y ds 2 43 1 2 19 2 6 ≤ ≤ ≤ ≤ t t t t ≤ ≤ ≤ ≤ 2 8 3 3 19 t 8 3 2 6 y x C 4 y ds 19 2 6 19 1 18. r t t i, 2i 6 8 x 0 t 2 j, 2 t i 2 j, 4 t j, 6 2 2 C3 (2, 2) C4 1 C2 4 y ds 0 4 t dt 2 2 6 C1 C1 x 1 2 x C2 4 y ds 4t 2 dt 4 16 2 3 2 82 x C3 4 y ds 4 8 6 t 42 dt x C4 4 y ds 6 48 t dt 16 2 3 16 2 3 2 82 16 2 3 i, r t 8 56 3 1 1 x C 4 y ds 2 4 2 19. C1: 0, 0, 0 to 1, 0, 0 : r t 1 ti, 0 ≤ t ≤ 1, r t 2t dt 0 z 2x C1 y2 z ds t 2 1 1 0 (1, 0, 1) C C2: 1, 0, 0 to 1, 0, 1 : r t 1 i 2 0 tk, 0 ≤ t ≤ 1, r t t dt tj t2 1 2t t2 2 1 0 k, r t 3 2 j, r t 4 3 1 (0, 0, 0) 1 (1, 0, 0) x 2x C2 y 2 z ds (1, 1, 1) y C3: 1, 0, 1 to 1, 1, 1 : r t 1 i 2 0 k, 0 ≤ t ≤ 1, r t 1 dt 3 2 4 3 t 23 6 t3 3 1 0 1 2x C3 y2 2x C z ds y2 Combining, z ds 376 Chapter 15 Vector Analysis tj, 0 ≤ t ≤ 1, r t 1 20. C1: 0, 0, 0 to 0, 1, 0 : r t 2x C1 j, r t 1 1 z y2 z ds 0 t 2 dt j 1 t3 3 1 0 1 3 k, r t 1 2 j t 2 dt k, r t 2 t3 3 t2 2 2 1 0 (0, 1, 1) C C2: 0, 1, 0 to 0, 1, 1 : r t 2x C2 tk, 0 ≤ t ≤ 1, r t t dt t 1 1 1 2 t 2 6 t2 2 1 0 1 1 x (0, 0, 0) (0, 1, 0) y y2 z ds 0 1 1 1 C3: 0, 1, 1 to 0, 0, 0 : r t 2x C3 tj t 1 3 2 t k, 0 ≤ t ≤ 1, r t 1 y2 2x C z ds 0 1 z ds 2 dt 0 t2 2 2 6 Combining, y2 5 6 21. x, y, z rt rt rt Mass C 12 x 2 3 cos t i 3 sin t i y2 z2 2 t k, 0 ≤ t ≤ 4 2k 2 3 sin t j 3 cos t j 2 3 sin t 3 cos t 4 2 2 13 3 sin t 2 x, y, z d s 0 1 3 cos t 2 4 2 2t 2 13 d t 4t 3 3 4 0 13 2 2 13 3 22. x, y, z rt rt rt Mass C 9 0 4t 2 dt 64 2 13 9t 2 4973.8 27 z 3 cos t i 3 sin t i 3 sin t 3 sin t j 3 cos t j 2 4 23. 2 t k, 0 ≤ t ≤ 4 2k 2 rt rt Mass cos t i sin t i sin t j, 0 ≤ t ≤ cos t j, r t x C 1 y ds x, y d s C 3 cos t 2 2 13 16 2 x, y, z d s 0 2t 13 d t 13 cos t 0 sin t dt cos t 0 sin t 1 24. rt rt Mass C 1 2 2t j 2j tk, 1 ≤ t ≤ 3 k, r t kz ds C 3 t2i 2t i 2t j, 0 ≤ t ≤ 1 2 j, r t x, y d s C 1 0 1 25. 4 rt rt Mass t2i 2t i 4t 2 3 y ds 4 3 2t 4 3t t 2 0 1 4t 2 5 x, y, z d s C 4t 2 1 32 0 12 4 dt 1 kt 4t 2 k 4t 2 12 k 41 41 12 5 5 dt 32 3 1 dt t2 22 1 27 1 S ection 15.2 26. rt rt rt Mass C Line Integrals 377 2 cos t i 2 sin t i 4 9 2 sin t j 2 cos t j 13 3 t k, 0 ≤ t ≤ 2 3k 27 . F x, y C: r t Ft xyi 4t i 4t 2 i 4i F dr j yj t j, 0 ≤ t ≤ 1 tj x, y, z d s C 2 k k 0 z ds 3t 3t 2 13 d t 2 2 0 2 rt 1 16 t 2 0 t dt 12 t 2 1 0 C 13 kt 13 2 k 28. F x, y C: r t Ft rt F C 16 3 t 3 35 6 6 xyi yj 4 sin t j, 0 ≤ t ≤ 4 sin t j 2 29. F x, y C: r t Ft rt 16 sin t cos t dt C 2 3x i 4yi 2 sin t j, 0 ≤ t ≤ 8 sin t j 2 cos t j 2 4 cos t i 16 sin t cos t i 4 sin t i dr 0 2 cos t i 6 cos t i 2 sin t i F dr 0 2 4 cos t j 2 64 sin2 t cos t 64 3 sin t 3 8 sin2 t 0 12 sin t cos t 2 16 sin t cos t d t 40 3 31. F x, y, z x2yi ti t4i i F C 2 sin2 t 0 2 30. F x, y C: r t Ft rt 3x i ti 3t i i F dr 4y j 4 t 2 j, 2≤t≤2 x zj x yz k C: r t Ft rt t2 j t 2j 2 k, 0 ≤ t ≤ 1 2 t 3k 4 4 t2j t j 4 t2 2 2tj 1 3t 2 4t dt C t2 2 2 0 2 dr 0 t4 t5 5 2t t 2t3 3 2 dt 1 2t 2 0 17 15 32. F x, y, z C: r t Ft rt x2 i 2 sin t i y2 j z2 k 12 t k, 0 ≤ t ≤ 2 14 tk 4 tk 8 cos2 t sin t t6 24 6 33. F x, y, z rt Ft dr 15 t dt 4 ti t2 i F C x2z i t2j ln t i 2tj dr 1 6yj yz2k 2 cos t j 4 cos2 t j 2 sin t j ln t k, 1 ≤ t ≤ 3 6t 2 j t 2 ln 2 t k 4 sin2 t i 2 cos t i F dr C 0 1 k dt t 3 8 sin2 t cos t 83 sin t 3 8 3 6 t 2 ln t 249.49 12 t 3 t ln t 2 dt 8 cos3 t 3 8 3 24 0 24 16 3 378 Chapter 15 Vector Analysis 34. F x, y, z rt Ft dr ti ti i F C xi yj zk x2 y2 z2 tj tj 2t 2 j dr 0 35. F x, y C: y x3 rt rt Ft xi 2y j e t k, 0 ≤ t ≤ 2 et k e2t et k dt 2 from 0, 0 to 2, 8 ti i ti t F C t 3 j, 0 ≤ t ≤ 2 3t 2 j 2t 3 j 6t 5 2 1 2t 2 e2t 2t e 2 t dt 6.91 F r Work dr 0 t 12 t 2 6t5 dt 2 t6 0 66 36. F x, y C: x rt rt Ft F r x2 i cos t, y cos3 t i 3 xyj sin3 t from 1, 0 to 0, 1 sin3 t j, 0 ≤ t ≤ 2 3 cos 2 t sin t i cos 6 t i 3 sin2 t cos t j cos3 t sin3 t j 3 cos4 t sin5 t sin4 t 1 cos2 t 2 cos2 t t sin t 2 3 cos8 t sin t 3 cos4 t sin t cos4 t 3 cos4 t sin t cos4 t 3 cos4 t sin t 2 cos4 t 6 Work C 1 cos8 t dr sin t 6 2 cos6 3 cos4 t sin t 6 cos6 t sin t 3 cos5 t 5 2 0 F 6 cos8 t sin t 0 3 cos4 t sin t dt 43 105 2 cos9 t 3 37. F x, y 2xi yj 6 cos7 t 7 C: counterclockwise around the triangle whose vertices are 0, 0 , 1, 0 , 1, 1 rt t i, i 3 t 1 j, ti 3 2 t i, r t F C1 0≤t≤1 1≤t≤2 t j, 2 ≤ t ≤ 3 i 1 On C1: F t Work On C2: F t Work dr 0 2t dt j t 1 1 2i t F C2 1 j, r t 2 dr 3 3 1 dt 1 2 i t 3 j t dt 3 2 On C3: F t Work 23 ti F dr t j, r t 23 2 C3 Total work C F dr 1 1 2 3 2 0 S ection 15.2 38. F x, y yi xj 4 x 2 Line Integrals 379 39. F x, y, z C: r t rt Ft F 4 cos 2 t cos 2 t d t 0 xi 2 cos t i yj 5z k 2 sin t j t k, 0 ≤ t ≤ 2 k 5t k C: counterclockwise along the semicircle y from 2, 0 to 2, 0 rt rt Ft F r 2 cos t i 2 sin t i 2 sin t i 4 sin2 t F C 2 sin t j, 2 cos t j 2 cos t j 4 cos2 t 4 0≤t≤ 2 sin t i 2 cos t i 5t 2 2 cos t j 2 sin t j r F C Work dr 0 5t dt 10 2 Work dr 2 sin 2 t 0 0 40. F x, y, z yz i xz j xyk 41. r t 0≤t≤1 F dr 3 sin t i 150k 3 cos t i 2 3 cos t j 10 t k, 0 ≤ t ≤ 2 2 C: line from 0, 0, 0 to 5, 3, 2 rt rt Ft F r 5t i 5i 6t 2 i 90 t 2 1 3t j 3j 2 t k, 2k 3 sin t j 1500 dt 2 10 k dt 2 1500 t 2 2 10 t 2 j 15t 2 k C F dr 0 1500 ft 0 lb Work C F dr 0 90 t 2 d t 30 42. r t rt F C ti i dr t 2 j, 0 ≤ t ≤ 1 2tj 10 5 34 16 3 x, y F x, y 24 46 11 0, 0 5i i 11 4 , 16 11 2, 4 39 4 , 16 1, 1 3j i i 5j 2j 11 3.5i i j 0.5j 2i i 4 2j j 1.5i i 44 rt F r 1.5j 6 5 4 43. F x, y (a) x2 i r1 t r1 t Ft F dr xyj 2t i 2i 4t i 3 2 t j 2t t 1 j, 1 ≤ t ≤ 3 (b) r2 t r2 t 23 2i 43 2 ti j t 2i 83 t 2 t j, 0 ≤ t ≤ 2 1j 1 dt 236 3 F C2 Ft dr 23 2 t2 23 tj t2 t dt 8t2 1 2t t C1 0 Both paths join 2, 0 and 6, 2 . The integrals are negatives of each other because the orientations are different. 44. F x, y (a) x2 y i r1 t r1 t Ft F C1 236 3 x y3 2 j t i t 2 1i 2tj 1 2t2 i t t 2 j, 0 ≤ t ≤ 2 t 1 t3j 2t 4 t 1 dt 256 5 dr 0 1 2t2 —CONTINUED— 380 Chapter 15 Vector Analysis 44. —CONTINUED— (b) r2 t r2 t Ft F C2 1 2 cos t i 2 sin t i 4 cos2 t j, 0 ≤ t ≤ 2 8 cos t sin t j 2 1 2 2 cos t 1 4 cos2 t i 2 1 2 cos t 8 cos3 t j 2 sin t 8 cos t sin t 1 2 cos t 8 cos3 t dt 256 5 dr 0 2 cos t 4 cos2 t Both paths join 1, 0 and 3, 4 . The integrals are negatives of each other because the orientations are different. 45. F x, y C: r t rt Ft F r F C yi ti i xj 2tj 2j 2ti 2t dr tj 2t 0. 0 46. F x, y C: r t rt Ft F r C 3y i ti i 3t 3 i 3t F 3 xj tj 3 3t 2 j tj 3t 3 0. 0 Thus, Thus, dr 47. F x, y C: r t rt Ft F r F C x3 ti i t3 t3 dr 2x 2 i t 2j 2tj 2t 2 i 2t 2 0. t x y j 2 48. F x, y C: r t rt xi yj 3 cos t j 3 sin t j 3 cos t j 9 sin t cos t 0 3 sin t i 3 cos t i 3 sin t i t2 j 2 t2 2 0 Ft F r 9 sin t cos t F C 2t t Thus, dr 0. Thus, 49. x 2t, y x C 10 t, 0 ≤ t ≤ 1 ⇒ y 10 5x or x y2 10 y , 0 ≤ y ≤ 10 5 10 3y 2 d y 0 y 5 3y 2 dy y3 0 1010 50. x x C 2 t, y 10 t, 0 ≤ t ≤ 1 ⇒ y 2 5x, 0 ≤ x ≤ 2 x2 2 2 3y 2 dx 0 x 75x 2 d x 25x 3 0 202 51. x 2t, y xy dx C 10 t, 0 ≤ t ≤ 1 ⇒ x 10 y , 0 ≤ y ≤ 10, dx 5 y3 75 y2 2 10 0 1 dy 5 190 OR 3 y dy 0 y2 25 y dy y 5x, d y xy dx C 5 d x, 0 ≤ x ≤ 2 2 y dy 0 5x 2 25x d x 5x 3 3 25x 2 2 2 0 190 3 S ection 15.2 52. x 2t, y 3y C Line Integrals 381 10t, 0 ≤ t ≤ 1 ⇒ y 2 5x, dy x dx 2 5 d x, 0 ≤ x ≤ 2 2 x dx y 2 dy 0 3 5x 7x2 5x 25 dx 0 14x 1084 3 125x2 dx 125 3 x 3 28 0 125 8 3 53. r t xt dx t i, 0 ≤ t ≤ 5 3 y t, y t d t, 2x C 0 0 5 2 1 x 1 2 3 4 5 dy y dx x 3y d y 0 2t dt 25 −1 −2 54. r t xt dx t j, 0 ≤ t ≤ 2 0, y t 0, dy 2x C y t dt 2 1 2 y dx x 3y d y 0 3t d t 32 t 2 2 6 0 −1 1 x 55. r t C1: t i, 3i xt dx 2x C1 t t, y t d t, dy 0≤t≤3 3 j, 3 ≤ t ≤ 6 0, y (3, 3) 3 2 0 1 C2 3 y dx 3, y t 0, dy t dt x 3 3y d y 0 2t dt 9 1 C1 x 2 3 C2: x t dx 2x C2 6 y dx y dx x x 3y d y 3 3 9 45 2 3t 63 2 3 dt 3t 2 2 6 6t 3 45 2 2x C 3y d y 56. r t C1: xt dx t j, t 3i 0, y t 0, dy y dx t 0≤t≤3 3 j, 3 ≤ t ≤ 5 t dt 3 −1 y x 1 2 3 C1 −2 2x C1 x 3y dy 0 3t d t 27 2 −3 C2 (2, − 3) C2: xt dx 2x C2 3, y t 0 3 d t, dy y dx y dx 5 5 x x 3y d y 3 2t 27 2 10 3 47 2 3 dt t 3 2 3t 3 10 2x C 3y d y 382 57. x t Chapter 15 t, y t 2x C Vector Analysis 1 t 2, 0 ≤ t ≤ 1, dx 1 dt, dy 2t 1 t2 t2 4t 2t dt t 1 dt 3 3t2 3t 4 2 2 t dt t3 3 1 y dx x 3y dy 0 1 6t3 0 2t2 t 0 11 6 58. x t t, y t 2x C t 3 2, 0 ≤ t ≤ 4, dx 4 dt, dy 2t 0 4 0 312 t dt 2 t3 2 y dx x 3y dy t 2 3t 3 2 31 t 2 33 t 2 2 dt 15 t 5 4 2 92 t 2 13 t 2 2t dt t2 0 96 1 32 5 16 592 5 59. x t dx t, y t d t, dy 2x C 2 t 2, 0 ≤ t ≤ 2 4t dt 2 y dx x 3y d y 0 2 2t 24t3 0 2t2 dt 2t2 t 2t dt 6t 2 4 t d t 6t 4 23 t 3 2 t2 0 316 3 60. x t dx 4 sin t, y t 4 cos t d t, dy 2x C 3 cos t, 0 ≤ t ≤ 3 sin t d t 2 2 y dx x 3y d y 0 2 8 sin t 3 cos t 4 cos t d t 12 cos 2 t 2 4 sin t 9 cos t 3 sin t d t 5 sin t cos t 0 12 sin2 t d t 52 sin t 2 61. f x, y h 12t 0 5 2 6 62. f x, y y C: line from 0, 0 to 3, 4 r rt rt 3t i 3i 5 4 t j, 0 ≤ t ≤ 1 4j C: line from 0, 0 to 4, 4 rt rt rt ti i 2 j t j, 0 ≤ t ≤ 4 Lateral surface area: 1 Lateral surface area: 4 f x, y d s C 0 5h d t 5h C f x, y d s 0 t 2 dt 82 S ection 15.2 63. f x, y C: x 2 xy y2 rt rt rt 1 1 from 1, 0 to 0, 1 cos t i sin t i sin t j, 0 ≤ t ≤ cos t j 2 64. f x, y C: x 2 x y2 rt rt rt 1 y 1 from 1, 0 to 0, 1 cos t i sin t i Line Integrals 383 sin t j, 0 ≤ t ≤ cos t j 2 Lateral surface area: 2 Lateral surface area: 2 f x, y d s C 0 cos t sin t d t C 2 0 f x, y d s 0 cos t sin t d t 2 sin2 t 2 65. f x, y C: y 1 rt rt rt h 1 2 sin t cos t 0 2 x 2 from 1, 0 to 0, 1 1 i 1 ti 21 41 1 tj t 2 1 t 2 j, 0 ≤ t ≤ 1 Lateral surface area: 1 f x, y ds C 0 h1 h 21 4 h 25 4 41 t ln 2 1 t 2 dt 1 41 5 t 2 ln 2 1 t 1 41 t 2 0 1.4789h 66. f x, y C: y 1 y 1 x from 1, 0 to 0, 1 2 rt rt rt 1 i 1 ti 21 41 1 tj t 2 1 t 2 j, 0 ≤ t ≤ 1 Lateral surface area: 1 f x, y d s C 0 2 1 1 1 t 41 t 2 1 t 2 dt 41 1 t 2 dt 1 0 1 2 0 t 2 1 t 1 ln 2 41 1 41 5 t t 2 dt 41 2 1 21 2 1 41 t 24 1 5 5 t 2 ln 2 1 t 2 t 2 0 1 1 21 64 1 25 2 23 32 5 ln 2 33 ln 2 64 1 ln 2 1 t 1 41 t 2 0 1 18 5 64 1 46 5 64 33 ln 2 5 2.3515 384 Chapter 15 xy 1 Vector Analysis 67. f x, y C: y x 2 from 1, 0 to 0, 1 cos t, then: You could parameterize the curve C as in Exercises 65 and 66. Alternatively, let x y 1 rt rt rt cos2 t sin2 t sin2 t j, 0 ≤ t ≤ 2 sin t cos t j 4 sin2 t cos2 t sin t 1 4 cos2 t 2 cos t i sin t i sin2 t Lateral surface area: 2 2 f x, y ds C 0 cos t sin2 t sin t 1 1 4 cos2 t 12 4 cos2 t dt 0 sin2 t 1 4 cos2 t 12 sin t cos t dt 1 4 cos2 t 3 2. Let u sin2 t and dv f x, y ds C sin t cos t, then du 2 32 0 2 sin t cos t dt and v 2 1 12 1 sin2 t 1 12 1 sin2 t 1 12 1 12 1 120 4 cos2 t 4 cos2 t 1 5 120 1 6 1 0 4 cos2 t 2 52 0 32 sin t cos t dt 32 1 1 120 4 cos2 t 11 52 1 25 5 120 0.3742 68. f x, y C: x2 x2 y2 y2 4 4 2 sin t j, 0 ≤ t ≤ 2 2 cos t j rt rt rt 2 cos t i 2 sin t i 2 Lateral surface area: 2 2π f x, y d s C 0 4 cos2 t 4 sin2 t 4 2 dt 8 0 1 cos 2t d t 8t 1 sin 2t 2 2 16 0 69. (a) f x, y rt rt rt S C 1 y2 2 sin t j, 0 ≤ t ≤ 2 2 cos t j (c) 5 4 z 2 cos t i 2 sin t i 2 2 −3 f x, y ds 0 1 2 4 sin2 t 2 dt x 3 3 y 2t (b) 0.2 12 4t 12 5 sin t cos t 0 12 37.70 cm2 7.54 cm3 S ection 15.2 1 x 4 Line Integrals 385 70. f x, y C: y rt rt rt 20 x 3 2, 0 ≤ x ≤ 40 ti i t 3 2 j, 0 ≤ t ≤ 40 3 12 tj 2 1 9 t 4 40 Lateral surface area: C f x, y d s 0 20 1 and dt 91 1 t 4 8 9u 1 d u. 9 t dt 4 Let u 40 1 1 t 4 9 4 t, then t 4 9 u2 20 0 1 9 t dt 4 20 1 12 u 9 91 1 1 u 8 u du 9 8 81 91 u4 1 179u 2 d u 8 u5 81 5 71. rt rt Ix C 179u3 3 850,304 91 1215 72. rt rt Ix C 7184 6670.12 a cos t i a sin t i y 2 x, y d s a sin t j, 0 ≤ t ≤ 2 a cos t j, r t 2 a cos t i a sin t i y 2 x, y d s a sin t j, 0 ≤ t ≤ 2 a cos t j, r t 2 a a a2 sin2 t 1 a dt 0 2 a2 sin2 t sin t a2 dt 0 2 a3 0 2 sin2 t dt a3 Iy C a4 0 2 sin3 t dt 0 Iy C x 2 x, y d s 0 a2 cos2 t 1 a dt 2 x 2 x, y d s 0 a 2 cos2 t sin t a2 dt 2 a3 0 cos2 t dt a3 a4 0 cos2 t sin t dt 0 73. S 25 60 50 40 30 20 10 3 x z 74. f x, y C: y S 8 y x 2 from 0, 0 to 2, 4 4 3 2 1 z Matches (b) Matches (c) 3 y 4 x 3 2 4 y (2, 4, 0) 75. (a) Graph of: r t For y 1 3 cos t i 3 sin t j 1 sin2 2 t k, 0 ≤ t ≤ 2 b and 3 3 2 1 z b constant, 3 sin t 1 1 1 1 b ⇒ sin t 2 sin2 2t 2 sin t cos t 4 sin2 t cos 2 t 4 sin2 t 1 42 b1 9 sin 2 t b2 . 9 x 3 4 3 4 y —CONTINUED— 386 Chapter 15 Vector Analysis 75. —CONTINUED— (b) Consider the portion of the surface in the first quadrant. The curve z 3 sin t j, 0 ≤ t ≤ 2. Hence, the total lateral surface area is 2 1 sin2 2t is over the curve r1 t 3 cos t i 4 C f x, y ds 4 0 1 sin2 2 t 3 d t 12 3 4 9 sq. cm. 4y3 2 (c) The cross sections parallel to the xz-plane are rectangles of height 1 3 1 y 2 9 and base 2 9 y 2. Hence, Volume 2 0 29 y2 1 4 y2 1 9 y2 9 dy 27 2 42.412 cm3. y 76. W C F 15 4 15xy dr C M dx 60 15x c 2cx d x N dy c y = c ) 1 − x 2) M N dx W x2y 15x 2 c cx2 cx2 d x, dy 1 −1 x 1 60 1 15x 2 c cx2 15 x c cx2 2 cx dx 120 w 16c 4c 4 8c 2 (parabola) 0⇒c 1 4 yields the minimum work, 119.5. Along the straight line path, y 0, the work is 120. 77. See the definition of Line Integral, page 1066. See Theorem 15.4. 79. The greater the height of the surface over the curve, the greater the lateral surface area. Hence, z 3 < z1 < z 2 < z 4 . y 4 3 2 78. See the definition, page 1070. 1 x 1 2 3 4 80. (a) Work 0 81. False 1 (b) Work is negative, since against force field. (c) Work is positive, since with force field. 82. False, the orientation of C does not affect the form f x, y d s. C C xy ds 2 0 t 2 dt 83. False, the orientations are different. 84. False. For example, see Exercise 32. S ection 15.3 85. F x, y rt rt Work y kt 1 k1 1 C 1 Conservative Vector Fields and Independence of Path y 387 xi ti 2t i F t 0 1 xy j t j, j dr kt 1 kt 1 2k 2t 3 ti t k1 kt 3 kt 2 kt 1 2t 2 −1 −1 0≤t≤1 2 1 (0, 1) (0, 0) 1 2 x tj kt 2 1 k1 2t i j dt t 0 1 t dt k 2t kt dt 3k 2t 2 0 k 12 k 12 Section 15.3 1. F x, y (a) r1 t r1 t Ft F C Conservative Vector Fields and Independence of Path xy j t 2 j, 0 ≤ t ≤ 1 2t j t3j 1 x2i ti i t2i dr (b) r2 r2 Ft 11 15 C sin i cos i sin2 dr 0 sin2 j, 0 ≤ 2 sin cos j sin3 j 2 ≤ 2 i t2 0 2t 4 dt F sin2 cos sin3 3 2 sin5 5 2 sin4 cos 2 0 d 11 15 2. F x, y (a) r1 t r1 t Ft F C x2 ti i t2 dr y2 i xj (b) r2 w r2 w Fw tj t2 t t2 2 1 2 3 F t dt 80 3 C t j, 0 ≤ t ≤ 4 1 j 2t ti 4 0 w2 i 2w i w4 dr 0 w j, 0 ≤ w ≤ 2 j w2 i 2 w2j w2 w3 3 2 0 2w w 4 w6 3 w4 2 w 2 dw 80 3 t3 3 t3 2 4 0 388 Chapter 15 yi xj Vector Analysis 3. F x, y (a) r1 r1 F F C sec i sec tan i dr 0 tan j, 0 ≤ sec2 j ≤ 3 tan i sec j 3 3 sec 3 tan2 sec3 ln sec d 0 sec 3 sec2 ln 2 1 3 sec3 d 1.317 sec d 0 tan 0 (b) r2 t r2 t Ft F C t 1 2t ti 1i 1 i t j, 1 j 2t 1j t 2t 1 t 3 0 0≤t≤3 t 3 dr 0 t1 dt 2t 1 12 2 1 2 dt 3 0 1 tt 1 ln 2 t 1 dt 1 2 t2 1.317 3 0 t2 3 t 1 14 14 dt 1 2 14 ln 1 2 1 2 t 0 7 1 ln 2 2 4. F x, y (a) r1 t r1 t Ft F C 23 1 ln 7 2 43 yi 2 i 3 dr x2 j ti j ti 3 3 t j, 0 ≤ t ≤ 3 2 3 t t 2j 2 3 t 2 dt 3 2 t 2 2 3 t 33 0 0 69 2 (b) r2 w r2 w Fw F C 2 1 i w 3 dr ln w i 1 j w ln w i e 1 3 ln w j, 1 ≤ w ≤ e3 2 ln w ln w 2 j 1 w 2 ln w 2 3 1 w dw 3 ln w 2 2 2 ln w 3 3 e3 1 69 2 5. F x, y N x Since e x sin y i e x cos y N x 1 i y 1 y2 ex cos y j M y e x cos y 6. F x, y N x Since 15x 2 y 2 i 30 x 2 y N x 10x 3 y j M y 30 x 2 y M , F is conservative. y x j y2 M y 1 y2 M , F is conservative. y 7. F x, y N x 8. F x, y, z curl F P y x z y ln z i x ln z j xy k z 0 so F is not conservative. x z N z N Since x M , F is not conservative. y S ection 15.3 9. F x, y, z curl F 11. F x, y (a) r1 t r1 t Ft F C Conservative Vector Fields and Independence of Path 10. F x, y, z curl F sin yz i xz cos yz j x y sin yz k 389 y2z i 2 xyz j xy 2k 0 ⇒ F is conservative. 2 xy i ti i 2t 3 i dr 0 0, so F is not conservative. x2j t 2 j, 0 ≤ t ≤ 1 2t j t2j 1 (b) r2 t r2 t Ft ti i 2t 4 i dr t 3 j, 0 ≤ t ≤ 1 3t 2 j t2j 1 4t 3 dt 1 C F 5t 4 d t 0 1 12. F x, y (a) r1 t r1 t Ft F C ye xy i ti i t dr j t xe xy j 3 j, 0 ≤ t ≤ 3 (b) F x, y is conservative since M y te 3t t2 3 3 e3t t2i te3t 3e 3t t2j t2 N x x ye xy e xy. e xy k. t 0 3 dt The potential function is f x, y e 3t 0 t2 3 e0 2 t dt e0 0 3 e 3t t2 0 13. F x, y (a) r1 t r1 t Ft F C yi ti i ti dr xj t j, j tj 0 C 0≤t≤1 (b) r2 t r2 t Ft F ti i t2i dr t 2 j, 2t j tj 1 0≤t≤1 (c) r3 t r3 t Ft ti i t3i dr t 3 j, 3t 2 j tj 1 0≤t≤1 t 2 dt 0 1 3 F C 2t 3 dt 0 1 2 14. F x, y (a) r1 t r1 t Ft F C xy2 i ti i 1 i t dr 2x2y j 1 j, t 1 j t2 2tj 3 1 1≤t≤3 (b) r2 t r2 t Ft t i 1 t 9 dr 1i 1 j 3 1t 2 1 t 3 3 j, 0≤t≤2 3 2i 1 t 9 1t 2 t 3 3 2 1 2 t 3j 1 2 1 dt t 3 F C 0 2 t 9 3 dt 2 t 3 dt ln t 1 ln 3 1 9 2 3t 3 0 7t 2 7t 3 3 7t 2 2 7t 3t 1 3t 4 94 0 44 27 390 Chapter 15 Vector Analysis 15. C y2 dx 2 xy d y xy2 k. Therefore, we N x 2y, F x, y y 2 i 2 xy j is conservative. The potential function is f x, y Since M y can use the Fundamental Theorem of Line Integrals. 4, 4 1, 0 (a) C y2 dx 2 xy d y x2 y 0, 0 64 y2 dx C (b) C y2 dx 2 xy d y x2 y 1, 0 0 (c) and (d) Since C is a closed curve, 2 xy d y 0. 16. C 2x 3y 1 dx x 3x y 5 dy 2x k. 2x C N Since M y f x, y x 2 3xy y2 2 3, F x, y x 5y 3y 1i 3x y 5 j is conservative. The potential function is (a) and (d) Since C is a closed curve, (b) C 3y x2 x2 1 dx 3xy 3 xy 3x y2 2 y2 2 y x x 5 dy 0, 1 0. 10 2x 2x C 3y 3y 1 dx 1 dx 3x 3x y y 5 dy 5 dy 5y 0, 1 2, e2 (c) 5y 0, 1 1 3 2 2e2 e4 17. C 2x y dx Since M F x, y y x2 N 2 xy i y2 dy x x2 2x, y 2 j is conservative. x2y x2 y y3 3 y3 3 y3 3 k. 0, 4 5, 0 0, 4 2, 0 18. C x2 Since M y2 dx y x2 2 xy d y N x y2 i 2y, 2 xy j F x, y The potential function is f x, y (a) C is conservative. The potential function is f x, y 64 3 64 3 (a) C x3 3 y2 dx xy2 2 xy d y k. x3 3 x3 3 8, 4 2x y dx x2 y2 dy x2 xy2 0, 0 0, 2 896 3 8 3 (b) C 2x y dx x2 y2 dy x2 y (b) C x2 y 2 dx 2xy dy xy 2 2, 0 19. F x, y, z yz i xz j xyk 20. F x, y, z i zj yk Since curl F 0, F x, y, z is conservative. The potential x yz k. function is f x, y, z (a) r1 t ti F C Since curl F 0, F x, y, z is conservative. The potential function is f x, y, z x yz k. (a) r1 t cos t i F C 2j dr tj t k, 0 ≤ t ≤ 4 4, 2, 4 sin t j x 2t i yz t 2 k, 0 ≤ t ≤ 1, 0, 2 x yz 0, 2, 0 32 (b) r2 t dr 1 2 1, 0, 0 (b) r 2 t t2i F C t 2 k, 0 ≤ t ≤ 2 4, 2, 4 2 t k, 0≤t≤1 1, 0, 2 dr x yz (0, 0, 0 32 C F dr x yz 1, 0, 0 2 S ection 15.3 21. F x, y, z 2y xi x2 zj 2y 4z k Conservative Vector Fields and Independence of Path 391 F x, y, z is not conservative. (a) r1 t r1 t Ft ti i 2t 2 t 2j 2t j ti 1 k, 0 ≤ t ≤ 1 t2 2t 3 2t 2 1j 2t 2 2 3 4k F C dr 0 t dt (b) r2 t r2 t Ft ti i 3t i F C tj j t2 2t 4 2t 2t 1 1 2 k, 0 ≤ t ≤ 1 1k 1 t2 2 j 2t 2t 1 2 4 2t 1 2 k 1 16 2t 3 dr 0 1 3t 17t 2 0 8t 2t 2 1 3 dt 5t 2 2 2t 6 1 3 1 4 0 5t 2t 1 16 2t 1 dt 17t 3 3 2 2t 1 17 6 22. F x, y, z yi xj 3xz 2 k F x, y, z is not conservative. (a) r1 t r1 t Ft F C cos t i sin t i sin t i dr sin t j cos t j cos t j t k, 0 ≤ t ≤ k 3t 2 cos t k cos 2 t 3t 2 cos t d t 0 sin2 t 0 1 t 3t 2 cos t d t 3t 2 sin t 6 sin t t cos t 0 t 0 3 t 2 sin t 0 6 0 t sin t d t 5 (b) r2 t r2 t Ft F C 1 2i 1 2t i k 2t j 1 t k, 0 ≤ t ≤ 1 3 3 0 22 t1 t1 2t k 1 dr 32 2t d t 3 3 0 t2 2t 3 dt 3 3 t3 3 t4 2 1 0 3 2 x sin z j x y cos x k 23. F x, y, z ez y i xj xyk 24. F x, y, z (a) r1 t r1 t Ft y sin z i t2i 2t i t4 F C F x, y, z is conservative. The potential function is f x, y, z x ye z k . (a) r1 t 4 cos t i F C t 2 j, 0 ≤ t ≤ 2 2tj t2 k 2 4 sin t j x ye z 4, 0, 3 3k , 0 ≤ t ≤ 0 cos dr 4, 0, 3 dr 4 8t i dr 0 dt 0 0 (b) r2 t F C 3k, 0 ≤ t ≤ 1 4, 0, 3 (b) r2 t r2 t Ft 4t i 4i 2 4 t j, 0 ≤ t ≤ 1 4j x ye z 4, 0, 3 0 16 t cos 4 t k 1 F C dr 0 0 dt 0 392 Chapter 15 Vector Analysis 3, 8 4, 3 25. C yi xj dr xy 0, 0 24 26. C 2x yi 2x yj dr x 49 y 2 2, 2 3 2, 2 27. C cos x sin y d x sin x cos y d y sin x sin y 0, 1 28. y dx x2 C 2x C x dy y2 arctan x y 2 3, 2 2 ,0 1, 1 3 1 x2 4 12 1, 5 29. C e x sin y d x e x cos y d y e x sin y 0, 0 0 30. x2 y2 2 dx 2y x2 y2 2 dy y2 7, 5 1 26 1 74 12 481 31. C y 2z d x x 3z d y 2x 3y d z xy 3yz 2 xz . F x, y, z is conservative and the potential function is f x, y, z 1, 1, 1 (a) xy 3 yz 2 xz 0, 0, 0 0, 0, 1 0 0 0 1, 1, 1 (b) xy 3yz 2 xz 0, 0, 0 1, 0, 0 xy 3 yz 2 xz 0, 0, 1 1, 1, 0 0 0 0 1, 1, 1 (c) xy 3yz 2 xz 0, 0, 0 xy 3yz 2xz 1, 0, 0 xy 3yz 2 xz 1, 1, 0 0 1 1 0 32. C zy dx xz dy x y dz x yz k, the integral is Note: Since F x, y, z y z i x z j x y k is conservative and the potential function is f x, y, z independent of path as illustrated below. 1, 1, 1 (a) x yz 0, 0, 0 0, 0, 1 1 1, 1, 1 (b) x yz 0, 0, 0 1, 0, 0 x yz 0, 0, 1 1, 1, 0 0 1 1 1, 1, 1 (c) x yz 0, 0, 0 x yz 1, 0, 0 x yz 1, 1, 0 0 0 1 1 2, 3, 4 33. C sin x d x z dy y dz cos x yz 0, 0, 0 12 1 11 4, 3, 1 34. C 6x d x 4z d y 4y 20z dz 3x 2 4yz 10z 2 0, 0, 0 46 35. F x, y Work 9x 2 y 2 i 3x 3 y 2 y 6x3y 5, 9 1 j is conservative. 30,366 36. F x, y Work 2x i y x2 y 1, 4 x2 j is conservative. y2 1 4 9 2 17 4 0, 0 3, 2 S ection 15.3 37. r t rt at Ft W C Conservative Vector Fields and Independence of Path 393 2 cos 2 t i 2 sin 2 t j 4 cos 2 t j 8 2 4 sin 2 t i 8 ma t F 2 cos 2 t i 1 at 32 dr C sin 2 t j 2 4 2 cos 2 t i cos 2 t i sin 2 t j sin 2 t j 4 sin 2 t i cos 2 t j d t 3 C 4 a3k 0 dt 0 38. F x, y, z a1i a2 j Since F x, y, z is conservative, the work done in moving a particle along any path from P to Q is Q q1, q2, q3 p1, p2, p3 \ f x, y, z a1x a1 q1 a2 y p1 a3z P a 2 q2 p2 a 3 q3 p3 F PQ . 39. F (a) r t dr 150 j ti i F C 50 j dt t j, 0 ≤ t ≤ 50 (b) r t dr ti i F C 1 50 1 25 50 50 6 0 t 2j t j dt 50 50 dr 0 150 d t 7500 ft lbs dr 50 t dt 7500 ft lbs 40. No. The force field is conservative. 41. See Theorem 15.5. 42. A line integral is independent of path if C F d r does not depend on the curve joining P and Q. See Theorem 15.6. 43. (a) The direct path along the line segment joining 4, 0 to 4, 4 and then to 3, 4 . 4, 0 to 3, 4 requires less work than the path going from (b) The closed curve given by the line segments joining 4, 0 , 4, 4 , 3, 4 , and 4, 0 satisfies C F dr 0. 44. No, the amount of fuel is not independent of the flight path. The vector field is not conservative. 45. Conservative. C F d r is independent of path. 46. Not conservative. The value of C F d r from 1, 0 to 1, 0 is positive if the path is above the x-axis, and negative if the path is below the x-axis. 47. False, it would be true if F were conservative. 49. True 48. True 50. False, the requirement is M y N x. 394 51. Let Chapter 15 Vector Analysis F Then 2f Mi M y Nj f y f i y 2f f j. x y 2 y 2f and M y N x N . x x f x 2f x2 . Since x 2 y2 0 we have Thus, F is conservative. Therefore, by Theorem 15.7, we have f dx y f dy x M dx C N dy C F dr 0 C for every closed curve in the plane. 52. Since the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing at a rate of 10 units per minute, then the potential energy is increasing at a rate of 10 units per minute. y x2 y x2 x2 x x2 x2 N x cos t i sin t i sin t i F C 53. F x, y (a) M M y N N x Thus, (c) r t F dr y2 y2 i x x2 y2 j sin t j, 0 ≤ t ≤ cos t j cos t j d t sin2 t 0 (b) r t y2 1 y 2y x2 y2 2 y2 y2 x2 M . y sin t j, 0 ≤ t ≤ cos t j cos t j d t sin2 t 0 cos t i sin t i x2 x2 y2 y2 2 F dr sin t i F C dr cos2 t d t t 0 1 y2 x 2x 2 x2 x2 y2 y2 2 (d) r t F dr cos t i sin t i sin t j, 0 ≤ t ≤ 2 cos t j cos t j d t 2 sin t i F C dr cos2 t dt dr 0 2 sin2 t t 2 0 cos2 t d t t 0 (e) arctan x y 1y i 1 x y2 y x2 y2 i x2 1 x x y2 j x y2 y2 j F This does not contradict Theorem 15.7 since F is not continuous at 0, 0 in R enclosed by curve C. Section 15.4 Green’s Theorem 395 Section 15.4 t i, 4i 12 16 y2 dx C Green’s Theorem 0≤t≤4 4≤t≤8 8 ≤ t ≤ 12 12 ≤ t ≤ 16 8 y 1. r t t 4 j, t i 4 j, t j, 4 (4, 4) 4 3 2 x2 dy 0 0 dt 12 t2 0 4 t dt 0 0 4 0 4 t 2 0 0 16 d t 16 1 x 1 2 3 4 16 8 12 2 16 12 t 2 0 0 dt 0 By Green’s Theorem, R 64 64 N x M dA y 0≤t≤4 4≤t≤8 8 ≤ t ≤ 12 8 4 4 2x 0 2y dy dx 0 8x 16 d x 0. y 2. r t t i, 4i 12 (4, 4) 4 3 2 t 4 j, ti 12 4 t j, y=x y2 dx C x2 dy 0 0 dt 12 t2 0 4 t 4 2 0 16 d t 128 3 64 3 64 . 3 1 x 1 2 3 4 12 8 t 2 dt 4 0 12 x t 2x 2 dt 0 64 4 By Green’s Theorem, R N x M dA y 2y d y d x 0 x2 dx 0 3. r t ti 8 y 2 dx C t 2 4 j, ti 8 x2 dy t j, 4 0 4 0 0≤t≤4 4≤t≤8 t4 dt 16 t4 16 t2 t3 dt 2 t dt 2 8 8 8 4 t 2 dt 224 5 8 128 3 t 2 dt 28 4 t 2 dt 32 15 By Green’s Theorem, N x M dA y 4 0 x 4 2x x 2 2y dy dx 0 x2 R 4 x3 2 x4 dx 16 32 . 15 y 4 3 2 1 (4, 4) C2 C1 x 1 2 3 4 396 4. r t Chapter 15 cos t i y2 dx C Vector Analysis sin t j, 0 ≤ t ≤ 2 2 y x2 dy 0 2 sin2 t cos3 t 0 2 sin t d t sin3 t d t sin2 t cos t cos2 t cos t d t −1 1 x 2 + y 2 =1 x 1 cos t 1 0 sin t 1 cos3 t 3 2 cos2 t d t 0 0 −1 sin t By Green’s Theorem, N x M dA y 1 1 2 0 sin3 t 3 1 1 1 x2 2x x2 2y dy dx 2 3 2 R 2 r cos 0 2r sin r dr d cos 0 sin d 2 0 3 0. 5. C: x 2 Let x y2 4 2 sin t, 0 ≤ t ≤ 2 . 2 2 cos t and y xey d x C ex dy 0 2 cos te2 sin t 2 2 4 4 x2 2 sin t e2 cos t 2 cos t d t 2 19.99 R N x M dA y ex x 2 xey d y d x 2 24 x2 ex xe 4 x2 xe 4 x2 dx 19.99 6. C: boundary of the region lying between the graphs of y 1 0 x and y xe x ex dx 3 x3 2.936 2.718 0.22 xe y d x C e x dy 0 xe x 1 0 x 3 3x 2e x d x 1 R N x M dA y N x M y 1 ex x3 xe y d y d x 0 xe x x3 e x d x 0.22 In Exercises 7-10, 1. 2 x y 7. C y x dx 2x y dy 0 2 x2 x dy dx 2 (2, 2) y=x 2x 0 x2 dx 1 4 3 8. Since C is an ellipse with a y C y = x2 − x x 1 2 2 and b y dy R 1, then R is an ellipse of area ab 1 dA Area of ellipse 2. 2 . Thus, Green’s Theorem yields x dx 2x S ection 15.4 9. From the accompanying figure, we see that R is the shaded region. Thus, Green’s Theorem yields y C y Green’s Theorem 397 (− 5, 3) (− 1, 1) (− 1, − 1) 4 2 (5, 3) (1, 1) x 2 4 x dx 2x y dy R 1 dA Area of R 6 10 56. 22 −2 (1, − 1) (− 5, − 3) − 4 (5, − 3) 10. R is the shaded region of the accompanying figure. y C y x dx 2x y dy R 1 dA Area of shaded region 1 25 9 8 2 4 2 1 −5 − 4 −3 −2 −1 −2 −3 −4 −5 x 12345 11. Since the curves y 2 xy d x C 0 and y x y dy 4 x 2 intersect at 1 2x d A 2, 0 and 2, 0 , Green’s Theorem yields 2 4 x2 1 20 2 2x d y d x 4 x2 R y 2 2 2 xy 8x 4x2 8 3 0 dx 2 x3 dx x4 2 32 . 3 2 2 4 2 x2 x3 3 4x 8 3 16 12. The given curves intersect at 0, 0 and 9, 3 . Thus, Green’s Theorem yields y2 dx C xy dy R 9 0 0 y x 2y d A 9 y dy dx 0 y2 2 x 9 dx 0 0 x dx 2 x2 4 9 0 81 . 4 13. Since R is the interior of the circle x 2 x2 C y2 2y a 2, Green’s Theorem yields 2y d A x2 a y2 dx 2 xy d y R a a a2 a2 4y dy dx x2 4 a 0 dx 0. 398 Chapter 15 Vector Analysis r sin , x 2xy d y R 2 1 0 2 cos 14. In this case, let y x2 C r cos . Then d A 2 r dr d and Green’s Theorem yields 1 0 cos y2 dx 4y dA 4 0 4 0 r sin r dr d r 2 sin dr d sin 0 4 3 1 cos 42 3 d 1 cos 3 0. 0 15. Since M y 2x x2 y2 N , x 16. Since M y N x 2ex sin 2y M dA y N we have x 0. we have path independence and N x M dA y R 0. R 17. By Green’s Theorem, sin x cos y d x C xy cos x sin y d y R 1 0 x y x sin x sin y 1 2 1 sin x sin y d A x 0 y dy dx x2 dx 1 x2 22 x3 3 1 0 1 . 12 18. By Green’s Theorem, e C x2 2 y dx e y2 2 x dy R 2 dA 2 Area of R 2 6 2 23 60 . 19. By Green’s Theorem, xy dx C x y dy R 2 0 1 3 x dA 2 1 1 r cos r dr d 0 4 26 cos 3 d 8. 20. By Green’s Theorem, (−1 , 1 ) y (1, 1) (2, 2) 3x 2 e y d x C ey dy R 2 1 2 3x 2e y d A 1 2 (−2 , 2 ) x 3x 2e y d y d x 2 1 2 2 11 3x 2e y d y d x 1 1 (− 2, − 2) (2, − 2) (1, − 1) 3x 2e y d y d x 2 1 2 3x 2e y d y d x 7 e2 e 2 (−1 , −1 ) 7 e2 16e 2 21. F x, y C: x 2 xy i y2 4 x yj e 2 2 2 e2 2e e 2e 1 e 2 16e 2e 1. 2 2 2 Work C xy dx x y dy R 1 x dA 0 0 1 r cos r dr d 0 2 8 cos 3 d 4 S ection 15.4 22. F x, y C: r Work C Green’s Theorem 399 ex 2 cos ex 3y i ey 6x j 3y d x ey 6x d y R 9 dA 9 since r 2 cos is a circle with a radius of one. 23. F x, y x3 2 3y i 6x 5 yj C: boundary of the triangle with vertices 0, 0 , 5, 0 , 0, 5 Work C x3 2 3y d x 6x 5 y dy R 9 dA 9 1 2 55 225 2 24. F x, y 3x 2 yi 4 xy 2 j x, y 1 dy dx 0 C: boundary of the region bounded by the graphs of y 9 x 0, x 9 9 2 Work C 3x 2 y dx 4 xy 2 d y 0 0 4y2 43 x 3 x1 2 dx 558 5 25. C: let x A 1 2 a cos t, y x dy C a sin t, 0 ≤ t ≤ 2 . By Theorem 15.9, we have y dx 1 2 2 a cos t a cos t 0 a sin t a sin t dt 1 2 2 a 2 dt 0 a2 t 2 2 a 2. 0 26. From the figure we see that C1: y C2: y C3: x A 1 2 1 2 3 3 x, dy d x, 0 ≤ x ≤ 2 2 2 1 x 4, dy dx 2 2 0, d x 2 0 0 2 4 3 y C2 x + 2y = 8 (2, 3) C1 3x − 2y = 0 x C3 2 1 0. 3 x dx 2 2 3 x 2 1 2 dx 0 0 2 1 2 3 4 1 x 2 x 2 4 dx 1 0 2 4 dx 2 4 27. From the accompanying figure we see that C1: y C2: y 2x 4 1, dy x 2, dy 2 dx 2 y (1, 3) x 4 2 x d x. −6 −4 −4 −6 Thus, by Theorem 15.9, we have A 1 2 1 2 1 2 1 x2 3 1 2x 1 dx 1 2 1 2 1 dx 3 1 2 x2 3 x 1 2x 4 x2 dx (− 3, − 5) 4 dx 4 dx 1 2 1 3 1 1 1 1 dx 3 x2 3 3 3 x2 dx 1 3x 2 x3 3 1 3 32 . 3 400 Chapter 15 Vector Analysis , 28. Since the loop of the folium is formed on the interval 0 ≤ t ≤ dx we have A 1 2 9 2 3t 0 31 t3 2t 3 dt and dy 12 3 2t t3 t4 d t, 12 t3 t5 t3 3 2t 1 t3 9 2 t4 12 3t 2 t3 31 1 t3 3 2 2t 3 12 dt 3t 2 t 3 1 2 0 t2 dt 13 0 t2 t 3 1 dt t3 1 3 dt 3 2 t3 1 0 0 3 . 2 1 x dy 2C 29. See Theorem 15.8, page 1089. 30. See Theorem 15.9: A y dx. 31. For the moment about the x-axis, Mx R y dA. Let N Mx 2A 1 2A 0 and M y 2 d x. C y 2 2. By Green’s Theorem, Mx C y2 dx 2 1 2 y 2 d x and y C For the moment about the y-axis, My R x dA. Let N My 2A 1 2A x 2 2 and M 0. By Green’s Theorem, My x2 dy C2 1 2 x 2 d y and x C x 2 dy. C 32. By Theorem 15.9 and the fact that x A 1 2 2 r cos , y r cos d r sin , we have r sin r sin d 1 2 y x dy y dx 1 2 x3 3 r cos r2 d . C 2 2 33. A 2 4 1 2A x2 dx 1 2A 4x 32 3 3 y = 4 − x2 x x2 dy C1 x2 dy C2 2 C1 For C1, dy x 1 2 32 3 2x dx and for C2, dy 2 0. Thus, 3 64 x4 2 2 2 1 x x2 2 2 x dx 0. −2 −1 C2 1 2 To calculate y, note that y y x, y 1 2 32 3 0, 8 5 2 0 along C2. Thus, x2 2 dx 3 64 2 4 2 16 2 8x 2 x 4 dx 3 16x 64 8x 3 3 x5 5 2 2 8 . 5 34. Since A Let x x y x, y area of semicircle a cos t, y 1 a2 1 a2 1 a2 , we have 2 2A a sin t, 0 ≤ t ≤ , then a 0 1 . Note that y a2 a 0 0 and dy 0 along the boundary y 0. a 2 cos 2 t a cos t d t 0 cos 3 t d t a 0 1 cos t sin2 t cos t dt cos 3 t 3 a 4a . 3 sin t sin3 t 3 0 0 a 2 sin2 t 0 a sin t d t sin3 t d t a 0 0, 4a 3 S ection 15.4 1 Green’s Theorem 401 35. Since A 0 x x3 dx x2 2 x4 4 1 0 1 1 , we have 4 2A 2. On C1 we have y x 3, dy 3x 2 d x and on C2 we have y x x, dy 2 C 1 d x. Thus, 2 C1 0 x2 dy x4 dx 0 x 2 3x 2 d x x2 dx 1 2 C2 x2 d x 8 15 1 y 6 y 2 2 6 5 2 3 (1, 1) C2 C1 x 1 y2 dx C 1 0 2 0 x6 dx 88 , 15 21 1 2a c 2 0, dy c b c b a a 2 1 x2 dx 2 7 2 3 8 . 21 x, y 36. Since A C1: y C2: y C3: y Thus, x y ac, we have 0 1 2A 1 , 2ac y (b, c) 2a C3 x x a , dy a , dy c b c b a a C2 dx d x. −a C1 x a 1 2ac 1 2ac x2 dy C 1 2ac 1 0 2ac a b 0 a b a a x2 c b c b 2 a a x dx b x2 c b a a b dx c a 2 1 0 2 ac x a 2 dx 2 abc 3 b 3 y2 dx C a c2 b a 2 dx b 1 c2 b a 2ac 3 x, y bc , 33 2 a 3 c . 3 37. A 1 2 a2 2 1 2 a2 1 0 2 cos 2 cos 2 d 1 2 a2 2 cos 2 2 1 0 1 0 d a2 3 22 a2 4 2 sin 1 sin 2 4 a2 4 2 0 a2 3 2 3 a2 2 38. A a 2 cos 2 3 d 0 cos 6 d 2 sin 6 6 ≤ . 0 Note: In this case R is enclosed by r a cos 3 where 0 ≤ 2 ≤ 3 d d 1 3 2 4 . Thus, 3 39. In this case the inner loop has domain A 1 2 1 2 4 2 4 2 3 ≤ 1 3 3 4 cos 4 cos 4 cos2 2 cos 2 4 3 3 3 3 4 sin sin 2 2 33 . 2 402 Chapter 15 Vector Analysis ≤ 2 and we let 1 1 u2 ,d u2 1 2 du . u2 2du 1 u2 1 u2 1 4 1 u2 1 6 arctan 3 3u 0 40. In this case, 0 ≤ u Now u A 1 sin , cos cos as ⇒ 1 2 2 ⇒ 2 9 cos and we have d 9 0 0 2 4 u2 u2 2 2 18 0 1 1 u2 du 3u 2 2 u 1 3u 2 1 3 du 3u 2 18 0 1 13 du 3u 2 18 0 1 23 du 3u 2 2 6 arctan 3 3u 0 12 1 32 3 3 0 6 32 6 u 3 1 3u 2 3 3 0 2 3. 0 41. I y dx x2 C x dy y2 y x2 y2 i x x2 N x y2 j. M y x2 x2 y2 . y2 2 (a) Let F F is conservative since F is defined and has continuous first partials everywhere except at the origin. If C is a circle (a closed path) that does not contain the origin, then F C dr C M dx N dy R N x M dA y 0. (b) Let r a cos t i a sin t j, 0 ≤ t ≤ 2 be a circle C1 oriented clockwise inside C (see figure). Introduce line segments C2 and C3 as illustrated in Example 6 of this section in the text. For the region inside C and outside C1, Green’s Theorem applies. Note that since C2 and C3 have opposite orientations, the line integrals over them cancel. Thus, C4 C1 C2 C C3 and F C4 dr F C1 dr C F dr 0. But, 2 F C1 dr 0 2 a sin t a 2 cos 2 t sin2 t a sin t a 2 sin2 t t a cos t a 2 cos2 t 2 a cos t dt a 2 sin2 t cos2 t d t dr 2. 2. 0 0 Finally, C F dr F C1 Note: If C were orientated clockwise, then the answer would have been 2 . y 3 2 C C2 x 4 C1 C3 −2 −3 S ection 15.4 42. (a) Let C be the line segment joining x1, y1 and x2, y2 . y dy y2 x2 y2 x2 y dx C Green’s Theorem 403 y1 x x1 y1 dx x1 x1 y`1 x2 x dy x1 y2 x2 y2 x2 y1 y1 x x1 y1 x1 x1 x2 y1 x1 x y2 x2 y1 x1 y1 x1 x2 y1 1 2 x2 dx x1 x1 x2 x1 y2 x2 y1 x1 y1 d x x1 x1 y2 y1 x y1 x2 1 2 x1 y2 x2 y1 x1 y dx x1 y2 x dy (b) Let C be the boundary of the region A Therefore, dA R 1 R 1 dA R d A. C 1 2 y dx C1 x dy C2 y dx x dy ... Cn y dx x dy where C1 is the line segment joining x1, y1 and x2, y2 , C2 is the line segment joining x2, y2 and x3, y3 , . . . , and Cn is the line segment joining xn, yn and x1, y1 . Thus, dA R 1 xy 2 12 x2 y1 x2 y3 x3 y2 ... xn 1 yn xn yn 1 xn y1 x1yn . 43. Pentagon: 0, 0 , 2, 0 , 3, 2 , 1, 4 , A 1 2 1, 1 1 4 1, 1 0 0 3 0 0 21 2 0 0 4 0 12 2 0 0 19 2 44. Hexagon: 0, 0 , 2, 0 , 3, 2 , 2, 4 , 0, 3 , A 1 2 0 0 4 0 N x 12 4 M dA y 6 y 45. C yn dx xn dy R 2a For the line integral, use the two paths C1: r1 x C2: r2 x yn dx C1 a y = a2 − x2 C2 x i, xi xn dy xn dy a≤x≤a a2 0 a2 a a a0 a2 x 2 j, x a to x a −a C1 x a yn dx C2 x2 x2 n2 xn 1 x a2 ny n 1 x2 dx R N x M dA y nxn dy dx (a) For n 1, 3, 5, 7, both integrals give 0. (b) For n even, you obtain n 2: 43 a 3 n 4: 16 5 a 15 n 6: 32 7 a 35 n 8: 256 9 a 315 (c) If n is odd then the integral equals 0. 404 Chapter 15 Vector Analysis 46. Since C F N ds R div F d A, then fg C f DN g ds C N ds f div R div f g d A R g f g dA R f 2g f g d A. 47. C f DN g gD N f d s C f DNg ds C gDN f d s f g dA R f R 2 g g 2 f g f dA R f 2 g g 2 f dA 48. C f x dx g y dy R x gy y fx dA R 0 0 dA 0 49. F N x F C Mi Nj M ⇒ y N x M dx C M y N dy 0 N x M dA y 0 dA R dr 0 R Section 15.5 1. r u, v z xy ui vj Parametric Surfaces uvk 2. r u, v x2 y2 u cos v i z2 u sin v j uk Matches (c) Matches (d) 4. r u, v x2 y 2 3. r u, v x2 y2 2 cos v cos u i z2 4 2 cos v sin u j 2 sin v k 4 cos u i 16 4 sin u j vk Matches (b) v k 2 Matches (a) 12 uk 2 12 x y2 8 5. r u, v y 2z ui 0 vj 6. r u, v z 2u cos v i y2 2 u sin v j 4u2 ⇒ z 12 2 u,x 2 Plane z 3 2 −4 345 5 x y Paraboloid z 4 4 x 4 y S ection 15.5 7. r u, v x2 z2 2 cos u i 4 vj 2 sin u k 8. r u, v x2 x2 z 3 Parametric Surfaces 3 cos v sin u j 9 cos2 v sin2 u sin2 v 1 5 405 3 cos v cos u i y2 y2 9 cos2 v cos2 u z2 25 z2 25 cos2 v 1 5 sin v k 9 cos2 v z Cylinder 9 x2 9 5 y y2 9 x 5 −3 Ellipsoid x 4 3 34 y For Exercises 9-12, r u, v u cos v i u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 . 5 z Eliminating the parameter yields z x2 y 2, 0 ≤ z ≤ 4. 2 x 2 y 9. s u, v z x2 u cos v i y2 u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 The paraboloid is reflected (inverted) through the xy-plane. 10. s u, v y x 2 u cos v i z2 u2 j u sin v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 The paraboloid opens along the y-axis instead of the z-axis. 11. s u, v u 2 k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 The height of the paraboloid is increased from 4 to 9. u cos v i u sin v j 4 u cos v i 2 12. s u, v z x 16 4 u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 y2 y2 64. It was x 2 14. r u, v y2 4. 4 cos v sin u j z 5 4 3 The paraboloid is “wider.” The top is now the circle x 2 13. r u, v 2u cos v i 2u sin v j u 4 k, z 3 2 1 2 cos v cos u i sin v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 z x2 16 y2 2 0≤u≤2 ,0≤v≤2 x2 4 y2 16 z2 1 1 −5 −4 −5 2 x 2 y 5 x 4 3 −3 −4 −5 5 y 15. r u, v 2 sinh u cos v i sinh u sin v j cosh u k, 9 6 z 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 z2 1 x2 4 y2 1 1 9 x 6 3 6 9 y 406 Chapter 15 2u cos v i Vector Analysis 2 u sin v j v k, z 16. r u, v 17. r u, v 0≤u≤ u sin u cos v i 1 cos u sin v j z 5 u k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 3 y tan z x −4 ,0≤v≤2 4 3 8 4 −2 2 x 4 2 4 y 3 x −4 −2 −3 −2 2 −3 −1 1 2 3 y 18. r u, v 0≤u≤ cos3 u cos v i 2 ,0≤v≤2 sin3 u sin v j z u k, 19. z y ui vj vk r u, v 2 −1 −1 1 1 y x 20. z 6 x ui y2 y vj 16 6 u vk 21. x 2 r u, v y2 16 4 cos u i 4 sin u j vk r u, v 22. 4x 2 r u, v x2 9 r u, v y2 4 23. z 4 sin u j vk x2 ui vj u2k 2 cos u i z2 1 r u, v 24. 1 2 cos v sin u j sin v k 25. z 4 inside x 2 v cos u i y2 9. v sin u j 4 k, 0 ≤ v ≤ 3 3 cos v cos u i r u, v 26. z x2 y 2 inside x 2 v cos u i y2 9. v 2 k, 0 ≤ v ≤ 3 27. Function: y x ,0≤x≤6 2 u cos v, z 2 u sin v 2 r u, v v sin u j Axis of revolution: x-axis x u, y 0 ≤ u ≤ 6, 0 ≤ v ≤ 2 28. Function: y x u, y u x3 2, 0 ≤ x ≤ 4 32 29. Function: x sin v x sin z, 0 ≤ z ≤ sin u sin v, z u Axis of revolution: x-axis cos v, z u3 2 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 30. Function: z x 4 4 y 2, 0 ≤ y ≤ 2 u, z 4 u 2 sin v Axis of revolution: z-axis sin u cos v, y 0≤u≤ ,0≤v≤2 Axis of revolution: y-axis u 2 cos v, y 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 S ection 15.5 31. r u, v ru u, v At 1, ru 0, 1 N u i vi u vj i 1. i i 1 1 y 2z 0 Direction numbers: 1, 1, Tangent plane: x x 1 y 2z j j 1 1 1 k k 0 1 2z i 1 j 0 2k At 1, 1, 1 , u ru 1, 1 i j v k, 1, k 1, 1 32. r u, v ru u, v ui i vj v 2 uv Parametric Surfaces uv k, 1, 1, 1 407 j, rv u, v 0 and v j, rv 0, 1 rv 0, 1 1 y k, rv u, v 1. j j 1, 1 , u i u k 2 uv 1 and v 1 k, rv 1, 1 2 rv 1, 1 ru 0, 1 1 k 2 1 2i 1 2j Tangent plane: x x N ru 1, 1 ijk 101 2 1 012 2 y 0 12 u k, 2 uk 1 2z k (The original plane!) 1 0 33. r u, v ru u, v rv u, v 2u cos v i 2 cos v i 2 u sin v i 3u sin v j 3 sin v j u 2 k, 0, 6, 4 2u k 34. r u, v ru u, v rv u, v At 2u cosh v i 2 cosh v i 2 u sinh v i 2 u sinh v j 2 sinh v j 2 u cosh v j 2 and v 0. 3u cos v j 2. 2 4i At 0, 6, 4 , u ru 2, N 2 ru 2, i 0 4 3j 2 and v 4 k , rv 2, rv 2, j 3 0 k 4 0 4, 0, 2 , u 2, 0 ru rv 2i ru N 2 k, rv 8i 8k 2, 0 4j 2 2 16 j 3 3z 12 4 0 12 k Direction numbers: 1, 0, 1 Tangent plane: x x 4 z z 2 2 0 Direction numbers: 0, 4, Tangent plane: 4 y 4y v j 2 6 3z 35. r u, v ru u, v 2ui v k , 0 ≤ u ≤ 2, 0 ≤ v ≤ 1 2 1 j 2 k 0 1 2 2 i, rv u, v i 2 0 2 2 1 k 2 j k ru ru A rv rv 1 0 j 0 1 2 2 du dv 0 22 408 Chapter 15 4 u cos v i 4 cos v i Vector Analysis 4 u sin v j 4 sin v j u2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 2u k 36. r u, v ru u, v rv u, v ru ru A 0 4 u sin v i 4 u cos v j rv rv 2 2 i j k 4 cos v 4 sin v 2u 4u sin v 4u cos v 0 64u 4 8u u2 0 8u 2 cos v i 4 128 2 3 8u 2 sin v j 16 u k 256u2 4 du dv 8u u 2 2 0 64 dv 3 128 3 22 1 37. r u, v ru u, v rv u, v ru ru A 0 0 a cos u i a sin u i k a sin u j a cos u j v k, 0 ≤ u ≤ 2 , 0 ≤ v ≤ b rv rv b 2 i j k a sin u a cos u 0 0 0 1 a a du dv 2 ab a cos u i a sin u j 38. r u, v ru u, v rv u, v ru ru A 0 a sin u cos v i a cos u cos v i a sin u sin v i a sin u sin v j a cos u sin v j a sin u cos v j a cos u k, 0 ≤ u ≤ π, 0 ≤ v ≤ 2 a sin u k rv rv 2 i j a cos u cos v a cos u sin v a sin u sin v a sin u cos v a2 sin u a2 sin u d u d v 0 k a sin u 0 a 2 sin2 u cos v i a 2 sin2 u sin v j a 2 sin u cos u k 4 a2 39. r u, v ru u, v rv u, v ru ru A 0 a u cos v i a cos v i a u sin v i a u sin v j a sin v j k u k, 0 ≤ u ≤ b, 0 ≤ v ≤ 2 a u cos v j a u cos v i a u sin v j a 2u k rv rv 2 b i j k a cos v a sin v 1 au sin v au cos v 0 au 1 a1 0 a2 a2 u du dv a b2 1 a2 S ection 15.5 40. r u, v ru u, v rv u, v ru rv a a b cos v cos u i b cos v sin u i a a b cos v sin u j b cos v cos u j b cos v k k 0 b cos v b cos v j b sin v a b cos v k Parametric Surfaces 409 b sin v k, a > b, 0 ≤ u ≤ 2 , 0 ≤ v ≤ 2 b sin v cos u i a b sin v sin u j a i b cos v sin u b sin v cos u j b cos v cos u b sin v sin u b sin u cos v a b cos u cos v a ru A 0 0 b cos v i rv 2 2 ba ba b cos v b cos v d u d v 4 2 ab 41. r u, v ru u, v rv u, v u cos v i cos v i 2u u sin v j sin v j 2u k u cos v j j sin v 2u u cos v u k, 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 u sin v i i cos v 2u u sin v k 1 0 u cos v i u sin v j 1 k 2 ru rv ru A rv 2 0 4 u u 0 1 4 1 du dv 4 uj j 6 17 17 1 36.177 42. r u, v ru u, v rv u, v ru ru A 0 sin u cos v i cos u cos v i sin u sin v i sin u cos v i sin u sin u 0 sin u sin v k, 0 ≤ u ≤ cos u sin v k sin u cos v k ,0≤v≤2 rv rv 2 cos u sin u j cos2 u sin u sin v k 1 1 cos2 u d u d v 22 ln 2 2 1 1 44. See the definition, page 1102. 43. See the definition, page 1098. 45. (a) From 10, 10, 0 (b) From 10, 10, 10 (c) From 0, 10, 0 (d) From 10, 0, 0 410 Chapter 15 Vector Analysis u cos v i ,0≤v≤ u sin v j from (b) 0, 0, 10 (c) 10, 10, 10 z −3 3 46. Graph of r u, v 0≤u≤ (a) 10, 0, 0 z 3 vk y y −3 3 3 3 −3 3 x x y 47. (a) r u, v 4 4 cos v cos u i cos v sin u j sin v k, (b) r u, v 4 4 2 cos v cos u i 2 cos v sin u j 2 sin v k, 0≤u≤2 ,0≤v≤2 z 4 −6 0 ≤ u ≤ 2 ,0 ≤ v ≤ 2 z −6 4 6 x −4 6 y x 6 6 y (c) r u, v 8 8 cos v cos u i cos v sin u j sin v k, (d) r u, v 8 8 3 cos v cos u i 3 cos v sin u j 3 sin v k, 0≤u≤2 ,0≤v≤2 z 9 0 ≤ u ≤ 2 ,0 ≤ v ≤ 2 z 12 3 3 x −9 y 12 x 12 − 12 y The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution. 48. r u, v (a) If u r 1, v x 2 2 u cos v i 1: 2 cos v i y 2 2 u sin v j v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 3 (b) If v 2 : 3 2 3 3x 2 3 2 −2 −1 1 2 y x 1 2 y 1 −1 −2 2 sin v j z 10 8 vk r u, y z 4 ui 3uj z 2 k 3 0≤z≤3 Helix 4 2 −2 x 2 2 −2 Line (c) If one parameter is held constant, the result is a curve in 3-space. S ection 15.5 49. r u, v ru rv ru rv 20 sin u cos v i 20 sin u sin v j 20 cos u k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 Parametric Surfaces 411 20 cos u cos v i 20 sin u sin v i 20 cos u sin v j 20 sin u cos v j 20 sin u k i j 20 cos u cos v 20 cos u sin v 20 sin u sin v 20 sin u cos v 400 sin2 u cos v i 400 sin2 u cos v i k 20 sin u 0 400 cos u sin u cos2 v cos u sin u k cos2 u sin2 u cos u sin u sin2 v k 400 sin2 u sin v j sin2 u sin v j sin u sin v 4 2 ru rv 400 sin u cos v 400 sin4 u 400 sin2 u 2 3 4 2 cos2 u sin2 u 400 sin u 2 3 2 S S dS 0 0 400 sin u d u d v 0 400 cos u 0 dv 0 200 d v 400 m2 50. x 2 Let x y2 z2 1 u sin v, and z sin v j u u2 1 u2 k 1 . Then, u cos v, y cos v i ru u, v rv u, v u sin v i u cos v j. 0. ru 1, 0 is undefined and rv 1, 0 2v k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 4π At 1, 0, 0 , u 51. r u, v ru u, v rv u, v ru ru A 0 0 1 and v j. The tangent plane at 1, 0, 0 is x 1. u cos v i cos v i u sin v j sin v j u cos v j z u sin v i 2k 2π rv rv 2 3 i j cos v sin v u sin v u cos v 4 4 u2 u2 d u d v k 0 2 2 sin v i 2 cos v j uk −4 −2 4 x 2 4 y 3 13 4 ln 3 2 13 52. r u, v ru u, v rv u, v ru rv ui i f u cos v j f u cos v j f u sin v j f u sin v k, a ≤ u ≤ b, 0 ≤ v ≤ 2 f u sin v k f u cos v k 53. Answers will vary. 54. Answers will vary. i j k 1 f u cos v f u sin v 0 f u sin v f u cos v fuf ui f u cos v j fu 1 2 f u sin v k ru A rv 2 0 fu b 1 fu fu 2 du dv a b 2 a fx 1 fx 2 dx since u x 412 Chapter 15 Vector Analysis Section 15.6 1. S: z 4 x S Surface Integrals z x 4 1, x z y 1 0 0 1 2 x, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 4 4 2y z dS 0 0 x 4 4 2y 4 0 2 dy dx 2 0 0 2y dy dx 2. S: z 15 x S 2x 2y 3y, 0 ≤ x ≤ 2, 0 ≤ y ≤ 4, 2 4 z x 2x 2, 3y z y 3, dS 1 4 9 dy dx 14 dy dx z dS 0 0 x 2 4 2y 15 15 x 14 dy dx 128 14 14 0 0 y dy dx 3. S: z 10, x 2 y 2 ≤ 1, z x 1 z y 1 1 1 0 x2 x S 2y z dS 1 2 0 2 0 x2 x 2y 10 1 0 2 0 2 dy dx r cos 0 2r sin 2 sin 3 10 r dr d 1 cos 3 5d 2 1 sin 3 2 cos 3 z x 5 0 10 4. S: z 2 32 x , 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 3 1 x x 1 2, 23 x 3 23 x 3 2 z y 1 0 x1 22 x S 2y z dS 0 1 0 0 x x 2y 0 2 dy dx x 0 1 2y 2 1 x dy dx 2 3 x5 0 2 x 1 dx 1 215 x 34 15 x 6 2 2 1 x 32 0 1 5 12 1 x3 0 2 1 x dx 1 1 x 32 0 51 12 3 x3 2 1 x 32 0 5 24 1 x1 0 2 1 x dx —CONTINUED— S ection 15.6 4. —CONTINUED— 2 3 2 18 2 18 2 18 2 18 52 18 5 24 1 Surface Integrals 413 5 24 x 1 x 0 x2 dx 2 0 1 2 1 2 1 dx 4 x 2 1 22 1 ln x 4 11 ln 42 61 2 288 5 ln 3 192 22 0.2536 1 2 1 51 24 2 53 48 2 15 2 96 z x 1 2 x x2 x2 x 0 13 ln 42 5 ln 192 3 z y 5. S: z 6 x 2y, (first octant) 6 3 0 6 x2 1, 2 2 5 y xy dS S 0 xy 1 xy 2 2 x9 0 3 0 x2 2 2 dy dx 4 3 2 1 y=3− 2x 6 0 dx 12 x dx 4 x4 16 6 0 1 x −1 1 2 3 4 5 6 6 2 6 3x 6 9x 2 2 2 x3 27 6 2 z x x4 0 6. S: z h, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 4 0 x2 4 x 2, 1 2 2 z y x 2 dx 0 1 2x 2 2 x4 4 2 dx dS S 0 xy dy dx 2 0 7. S: z z x 9 x 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ x, 2x, z y 2 0 2 xy dS S 0 y xy 1 4x 2 dx dy 391 17 240 1 y, 2 z y 1 x 2 3904 15 1 8. S: z 1 xy, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 2 4 4 z x xy dS S 0 0 xy 1 y2 4 x2 dy dx 4 160 5 3 9. S: z 10 x2 S x2 y 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 2 2 2xy dS 0 0 x2 2xy 1 4x 2 4y 2 dy dx 11.47 414 Chapter 15 Vector Analysis x 2 2 10. S: z cos x, 0 ≤ x ≤ x2 S 2 ,0≤y≤ 2 x2 2xy dS 0 0 x2 2xy 1 sin2 x dy dx 0 x3 4 1 sin2 x dx 0.52 11. S: 2x x, y, z m 3y x2 6z y2 x2 12 (first octant) ⇒ z 2 1 x 3 2 1 y 2 5 4 3 2 1 y y2 2x 3 1 x2 R 1 3 y 2 dy dx 2 1 2 y=4− 2 x 3 dA R x 1 2 3 4 5 6 7 6 7 6 12. S: z x, y, z m S 6 0 6 4 0 −1 x2 4 0 2 x 3 y2 1 4 3 2 x 3 3 dx 743 x 63 14 x 6 1 4 8 2 x 3 46 0 364 3 z a2 kz x2 a kz dS R k a2 k a2 R x2 x2 ka R y2 y2 dA 1 a x2 a2 y2 x x2 dA 2 y2 a2 y x2 2 y2 dA a x a y a2 ka dA R ka 2 a2 2 ka3 13. S: r u, v ru ru ru i, rv rv rv y S ui j 1 j 2 vj 1 k 2 k v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 2 14. S: r u, v 2 cos u i 2 sin u j v k, 0 ≤ u ≤ 2 0≤v≤2 ru rv x S 2 , 2 cos u i 2 2 sin u j 2 y dS 0 0 2 cos u 2 sin u 2 du dv 16 1 j 2 2 k 1 5 2 v 5 5 du dv 2 65 5 dS 0 0 15. S: r u, v ru rv ru ru k rv rv xy dS S 2 cos u i 2 sin u j v k, 0 ≤ u ≤ 2 ,0≤v≤2 2 sin u i 2 cos u j 2 cos u i 2 cos u i 2 0 0 2 2 sin u j 2 sin u j 2 8 8 cos u sin u du dv S ection 15.6 16. S: r u, v ru rv x S Surface Integrals 415 4 u cos v i 12 u cos v i 4 u sin v j 3u k, 0 ≤ u ≤ 4, 0 ≤ v ≤ 16u k 20u 10,240 3 12 u sin v j 4 y dS 0 0 4u cos v 4u sin v 20u du dv 17. f x, y, z S: z x x2 2, x 2 y2 z2 y2 ≤ 1 1 1 1 2 1 x2 f x, y, z dS S 1 x2 x2 r2 0 2 0 1 y2 x 2 2 2 1 1 2 0 2 dy dx 2 r cos r 2 cos2 r4 cos2 4 11 4 2 r dr d 2 0 2 0 r2 r4 4 9 4 1 8 4r cos 4r 3 cos 3 4 r dr d 1 2 0 2 2r 2 0 d 2 0 cos 2 2 4 cos 3 4 sin 3 2 d 2 0 2 9 4 1 sin 2 2 18 4 4 19 2 4 18. f x, y, z S: z x2 xy z y 2, 4 ≤ x 2 y 2 ≤ 16 xy S 2 0 2 4 2 f x, y, z dS S x2 4 y2 r1 1 4x 2 4y 2 dy dx 0 2 r 2 sin r2 cos 4 1 4r 2 r dr d 4r 2 sin cos dr d 0 2 2 1 1 12 4r2 32 2 sin cos d 65 65 17 17 sin2 12 2 0 0 19. f x, y, z S: z x2 x2 y2 z2 y2 ≤ 4 2 4 4 2 x2 y 2, x 2 f x, y, z dS S 2 x2 x2 4 4 x2 y2 x2 y2 2 1 x x2 y 2 dy dx y2 2 y x2 y2 2 dy dx 2 2 x2 2 x2 4 4 2 x2 y2 x2 y2 x2 x2 y2 2 2 2 x2 x2 r 2 dr d 0 0 y 2 dy dx 2 2 2 0 r3 3 2 d 0 16 3 2 0 32 3 416 Chapter 15 x2 x2 Vector Analysis y2 z2 1 2 20. f x, y, z S: z y 2, x y2 ≤ 1 x2 y2 y2 y 2 dy dx 16 3 x2 2 x2 x2 2 0 0 f x, y, z dS S S y2 2 1 x x 2 2 y x 2 2 y 2 y 2 dy dx 2 x2 S y2 dy dx y2 2 cos 2 S x2 r 2 dr d sin2 16 3 cos3 d 0 1 0 cos d 16 sin 3 21. f x, y, z S: x 2 y2 x2 y2 z2 sin3 3 0 0 9, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9 9 y2 3 9 dy 2 y y 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9. z2 1 3 Project the solid onto the yz-plane; x 3 9 f x, y, z dS S 0 3 0 0 9 9 y2 z2 3 9 y y 9 3 9 y 3 3 2 y y2 2 0 2 dz dy z3 3 0 9 9 0 3 dz dy 2 9z dy 0 0 324 0 972 arcsin 972 0 2 486 22. f x, y, z S: x 2 y2 x2 y2 z2 x 2. z2 1 x 9 3 2 9, 0 ≤ x ≤ 3, 0 ≤ z ≤ x 9 x2 Project the solid onto the xz-plane; y 3 x f x, y, z dS S 0 0 x2 9 x2 0 2 dz dx 3 0 3 0 x 9 0 z2 3 3 9 9x x 2 dz dx 0 3 3 9 x x2 9 3 3 2 9z 12 z3 3 dx x dx 0 3 9 x2 27 9 x2 12 x3 dx 3 27x 9 0 x3 9 0 x2 12 dx Let u x 2, dv x9 dx, then du 3 2x dx, v x2 9 x2 0 x 2. 2x 9 x 2 dx x2 0 3 0 81 2 9 3 x2 32 0 81 18 99 S ection 15.6 23. F x, y, z S: x y 3z i z x i N dS R Surface Integrals 417 4j yk 1 y 1 (first octant) y j z k 1 1 0 1 0 1 0 1 x x x x G x, y, z G x, y, z F S 1 y = −x + 1 R x 1 F G dA 0 1 0 1 0 1 3z 4 y dy dx 31 x y 4 y dy dx 1 3x y2 2y dy dx y 0 1 3xy dx 0 2 1 0 1 x 3x 1 4 3 x 1 x dx 2 0 2x2 dx 24. F x, y, z S: 2x G x, y, z G x, y, z F S xi 3y z 2x 2i N dS yj 6 (first octant) 3y 3j z k 1 3 y 6 2 y = − 2x + 2 3 3 2x 3 0 2 R F R G dA 0 3 0 2x 42 x 3 43 x 9 4x 2x 2 3 4 3y dy dx 3 2 2 x 3 2 x 3 2 0 2 x 1 2 3 2 33 dx 12 25. F x, y, z S: z 9 xi x2 x2 2xi N dS yj zk 4 2 y y 2, 0 ≤ z y2 z 2y j F R x2 + y2 ≤ 9 R x 2 4 G x, y, z G x, y, z F S 9 k G dA R −4 −2 −2 2x 2 2x 2 R 2y 2 2y 2 y2 z dA x2 y 2 dA −4 9 x2 R 2 0 2 0 3 9 dA r2 0 9 r dr d 9r 2 2 3 r4 4 d 0 243 2 418 Chapter 15 xi y2 36 z2 x2 z 36 yj Vector Analysis zk 6 5 y 26. F x, y, z S: x 2 z 36 (first octant) y2 36 x x2 y2 F R x2 + y2 = 62 4 G x, y, z G x, y, z F G F S x2 y 2 y2 i 36 y2 36 x 2 G dA R 2 0 3 2 R y x2 y2 36 6 0 y 2 j k 36 36 x 2 y2 dA 1 x 1 2 3 4 5 6 x2 36 x 2 N dS z 36 x2 y2 36 r dr d 36 r 2 (improper) 108 27. F x, y, z S: z x2 4i 3j 5k y2 ≤ 4 y2 z 2y j F R 1 y x 2 + y2 ≤ 4 y 2, x 2 x2 2xi N dS G x, y, z G x, y, z F S R x k G dA R 2 0 2 0 2 0 2 −1 −1 1 8x 6y 8r cos 5 dA 6r sin 2r 3 sin 16 sin 5 r dr d 52 r 2 10 d 2 2 0 83 r cos 3 64 cos 3 64 sin 3 d 0 16 cos 10 0 20 28. F x, y, z S: z G x, y, z G x, y, z F G F S xi a2 z yj x2 2z k y2 a y x 2 + y2 ≤ a 2 a2 a2 x x2 x2 y y2 F y2 i 2 a2 a2 G dA R 2 0 2 y x2 y2 y j 2 k x2 2a 2 dA y2 y2 3x 2 3y 2 a2 x 2 2a 2 y2 −a a x a2 N dS x2 x2 y2 x2 −a 2 a2 R 3x 2 3y 2 a2 x 2 a 0 3r 2 2a 2 r dr d a2 r2 a 3 0 2 0 r3 a 2 2 a 0 a 32 0 r dr d 2 r2 2 2a 2 0 r a d 2 r2 dr d 2 a 3 0 2 r2 a2 23 ad 3 2a 2 22 a 3 ad 0 r2 2a 2 0 a2 r2 0 d 3 0 0 S ection 15.6 29. F x, y, z 4xy i z2 j yz k 0, x 1, y 0, y 1, z 0, z 1 1 z Surface Integrals 419 S: unit cube bounded by x S1: The top of the cube N k, z F S1 1 1 1 N dS 0 0 y 1 dy dx 1 2 1 x 1 y S2: The bottom of the cube N F S2 S3: The front of the cube N i, x F S3 k, z N dS 0 1 0 1 1 1 1 y 0 dy dx 0 0 N dS 0 0 4 1 y dy dz 2 S4: The back of the cube N i, x F S4 S5: The right side of the cube N 1 1 0 N dS 0 0 j, y F S5 1 1 1 4 0 y dy dx 0 N dS 0 0 z 2 dz dx 1 3 S6: The left side of the cube N j, y F S6 0 1 1 N dS 0 0 z 2 dz dx 1 3 Therefore, F S N dS 1 2 x yi 0 2 0 1 3 1 3 5 . 2 30. F x, y, z S: z 1 yj 0 1 k zk x2 z y 2, z x2 y2 2y j y G x, y, z G x, y, z F G F S 2x i 2x x N dS 2y y F R 1 G dA x2 y2 x2 xx 2 2xy 2xy y2 y2 1 dA 1 R 2 0 2 0 1 r2 0 2r 2 cos 1 sin 2 sin 1 r dr d 3 4 sin2 4 2 0 3 4 cos d 3 2 The flux across the bottom z 0 is zero. g x, y , is defined as 31. The surface integral of f over a surface S, where S is given by z n f x, y, z dS S lim →0 i f xi , yi , z i 1 Si . (page 1108) See Theorem 15.10, page 1108. 32. A surface is orientable if a unit normal vector N can be defined at every nonboundary point of S in such a way that the normal vectors vary continuously over the surface S. 420 Chapter 15 Vector Analysis 34. Orientable 33. See the definition, page 1114. See Theorem 15.11, page 1114. 35. E S: z E S yz i 1 xz j x 2 xyk y2 E R N dS gx x, y i xz j x yk gy x, y j k dA x x2 y x2 1 1 1 yz i R 1 x y dA y2 i 1 y2 x2 j k dA R 2 xyz 1 x2 y2 3xy dA R 1 x2 3xy dy dx 0 36. E S: z xi yj 1 E 2z k x2 y2 g x, y E R N dS gx x, y i yj x2 x2 y2 2z k gy x, y j x x2 y2 x2 dA k dA i y x2 j k dA S xi R 1 1 x2 y2 y2 y2 y2 1 2z dA y2 R 1 x2 y2 R 21 1 x2 R 2 0 2 x2 1 x2 1 0 y2 dA y2 2 r2 r dr d 1 r2 8 3 37. z m S x2 y 2, 0 ≤ z ≤ a k dS k R 1 x x k x2 2 2 y x 2 2 y y2 2 y 2 dA k R 2 dA 2 k a2 Iz S k x2 2 a y 2 dS R 2 dA 2k 0 0 r 3 dr d a2 2 2ka 4 2 4 a 2m 2 2k a 4 2 2k a 2 S ection 15.6 38. x 2 z y2 ± Surface Integrals 421 z2 a2 2 S a2 x2 k dS y2 2k R m 1 a x2 a a2 y2 2 0 x x2 dA 2 y2 2 a 0 a2 r a2 y x2 r2 2 y2 dr d dA 2k R a2 a2 2ka 0 2ka k x2 S r2 4 ka 2 Iz 2 y 2 dS y2 a x2 22 a 3 2 a 0 2k R x2 a2 r2 y dA 2 a 32 0 2ka 0 r3 a2 r2 dr d (use integration by parts) 2ka r2 a2 23 a2 3 r a2 r2 2 2ka Let u 39. x 2 y2 22 a 4 ka2 3 r2 12 22 am 3 2r dr, v a2 r 2. r 2, dv dr, du a 2, 0 ≤ z ≤ h 1 a2 x2 h z x, y, z y ± Project the solid onto the xz-plane. Iz 4 S h a x2 x2 0 0 h a 0 y 2 1 dS x a a y 4 4a 3 0 h a2 x2 1 x a2 x2 2 0 2 dx dz 1 dx dz a2 x 2 x a a 4a 3 0 arcsin dz 0 4a 3 2 h 2 a 3h 40. z x2 y 2, 0 ≤ z ≤ h z Project the solid onto the xy-plane. h Iz S h x2 h h h y 2 1 dS x2 y x2 h 2 0 0 x2 y2 1 4x 2 4y 2 dy dx x r2 1 2 1 h 1 12 4h 60 32 4r 2 r dr d 32 4h 1 1 120 1 4h 4h 52 2 120 60 1 4h 32 10h 60 6h 1 1 422 Chapter 15 16 x2 0.5z k F N dS S Vector Analysis y 2, z ≥ 0 41. S: z F x, y, z F R gx x, y i gy x, y j 0.5 16 k dA R 0.5 z k 2x i 2y j k dA 0.5 z dA R 2 4 R x2 y 2 dA 2 0.5 0 0 16 r 2 r dr d 0.5 0 64 d 64 42. S: z F x, y, z 16 x2 y2 0.5z k F N dS F R gx x, y i gy x, y j x x2 0.5 i k dA y x2 j k dA S 0.5 z k R 16 y2 16 16 x2 2 y2 0.5 z dA R 2 4 R y2 d 64 d 3 64 3 (c) r u, 0 4 cos 2u i 4 sin 2u j 0.5 0 0 16 r 2r dr d 0.5 0 43. (a) 4 −6 z −6 6 −4 6 y (b) If a normal vector at a point P on the surface is moved around the Möbius strip once, it will point in the opposite direction. This is a circle. z 4 x −2 2 x 2 y −4 (d) (construction) (e) You obtain a strip with a double twist and twice as long as the original Möbius strip. Section 15.7 Divergence Theorem 423 Section 15.7 Divergence Theorem 1 dA. 1. Surface Integral: There are six surfaces to the cube, each with dS z 0, N k, F N z 2, S1 a a 0 dA a 2 dA S2 0 a 2 dx dy 0 0 z a, N k, F N z 2, a4 x 0, N i, F N 2x, S3 0 dA 0 a a x a, N i, F N 2x, S4 2a dy dz 0 0 2a dy dz 2a3 y 0, N j, F N 2y, S5 0 dA 0 a a y a, N F j, N dS F a4 N 2a 3 2y, S6 2a dA 0 0 2a dz dx 2a 3 Therefore, s 2a 3 a 4. Divergence Theorem: Since div F a a 0 a 2z, the Divergence Theorem yields a a div F dV 0 Q 0 2z dz dy dx 0 0 a 2 dy dx a 4. 2. Surface Integral: There are three surfaces to the cylinder. Bottom: z 0 dS S1 0, N 0 k, F k, F N z2 h z Top: z h, N h 2 dS N z2 4 h2 x 2 2 y h 2 Area of circle) 2 cos u i 2 sin u j 2 cos u j, rv 2 sin u j 8 sin2 u 8 cos u 0 0 2 S2 Side: r u, v ru ru F ru F S3 v k, 0 ≤ u ≤ 2 , 0 ≤ v ≤ h k 2 sin u i rv rv N dS F S 2 cos u i 8 cos2 u h 2 8 sin u du dv 0 2 2z 4 h2. 2z 4 h 2. 2 0 Therefore, N dS 0 4 h2 2 Divergence Theorem: div F 2 2 0 h 2z dV 0 Q 0 2zr dz dr d 424 Chapter 15 Vector Analysis z 6 3. Surface Integral: There are four surfaces to this solid. z 0, N 0 dS S1 k, F 0 j, F 6 N z y 0, N z dS S2 N 6 0 z 2y z, dS 6 dA z2 0 dx dz 6 3 y z dx dz 0 6z dz dz dy 2y 2 dy 2x 5y 6 x 18. 9 36 x x 0, N y dS S3 i, F 3 0 N 6 0 2y y 2x, dS 3 dA 6y y dz dy 0 x 2y z 2x 6, N 5y F i 2j k ,F 6 3 6 0 N 2y 3z , dS 6 dA 3 3z dz dy 0 18 36 9 45 11y dx dy 0 90 90y 20y 2 dy 45 S4 Therefore, s N dS 0 Divergence Theorem: Since div F dV Q 1, we have 1 Area of base 3 Height 1 96 3 18. Volume of solid 4. F x, y, z xy i zj x yk 4, z 4 x and the coordinate planes S: surface bounded by the planes y Surface Integral: There are five surfaces to this solid. z 0, N x S1 k, F y dS N 4 0 4 x x 0 y 4 y dy dx 0 4x 8 dx 64 y 0, N z dS S2 j, F 4 0 N 4 0 x z 4 z dz dx 0 4 2 x 2 dx 32 3 y 4, N z dS S3 j, F 4 0 N 4 0 x z 4 0 z dz dx i, F N 4 4 4 2 x 2 dx 32 3 x 0, N xy 0 dS 0 1 xy 2 4 4 xy dS S4 0 0 x z 4, N 1 xy 2 S i k ,F 2 x y N x y , dS 2 dA 2 dA 0 0 xy 32 3 32 3 x 0 y dy dx 128 64. 128 S5 Therefore, F N dS 64 Divergence Theorem: Since div F 4 4 0 4 0 x y, we have 64. div F d V 0 Q y dz dy dx S ection 15.7 5. Since div F 2x 2y a Divergence Theorem 425 2z, we have a 0 a a div F dV 0 Q a 0 0 2x 2ax 0 2y 2ay 2z dz dy dx a a a 2 dy dx 0 2a 2x 2a 3 dx a 2x 2 2a 3x 0 3a 4. 6. Since div F 2 xz 2 2 a 3xy we have a 0 a a a 0 div F dV 0 Q a 0 0 2xz2 24 xa 3 2a3 2 3xy dz dy dx 0 23 xa 3 2a 3xya dy dx 2a2 35 a. 4 2 xyz 33 xa dx 2 16 a 3 7. Since div F 2x 2x 2 xyz a 2 0 0 2 div F dV Q Q a 0 a 0 2 0 2 0 2 xyz dV 0 2 2 sin cos sin sin cos 2 sin ddd 2 0 5 sin cos sin3 a cos 5 ddd 2 1 2 5 sin cos d d 0 sin2 2 2 d 0 0. 8. Since div F y z y a z, we have a2 x2 a2 x2 0 a2 x2 y2 2 a 0 0 a2 r2 div F dV a Q 2 0 a 0 z dz dy dx 0 zr dz dr d a 2r 2 r3 dr d 2 2 0 a 2r 2 4 r4 8 a 2 d 0 0 a4 d 8 a4 . 4 9. Since div F 3 dV Q 3, we have 3 Volume of sphere 3 4 3 2 3 32 . 10. Since div F xz, we have 4 3 3 9 9 y2 4 3 3 xz dV 0 Q y2 xz dx dy dz 0 z 0 dy dz 2 0. 11. Since div F 1 2y 4 3 3 1 9 2y, we have y2 4 3 4 2y dV 0 Q 9 y2 2y dx dy dz 0 3 4y 9 y 2 dy dz 0 4 9 3 3 y2 32 3 dz 0. 426 Chapter 15 y2 y2 Vector Analysis x2 ez, we have 16 256 256 16 0 8 x2 x2 8 12. Since div F x2 Q ez dV 0 2 0 2 0 12 x2 y2 x2 r2 r2 y2 2 ez dz dy dx 16 ez r dz dr d 0 0 8r 3 262,104 5 re8 14 r 2 re r 2 dr d 131,052 5 100e8 d 200e8 . 13. Since div F 3x 2 x2 6 4 0 0 4 0 y 4x 2, we have 6 0 4 6 4 x2 dV 0 Q 4x 2 dz dy dx 0 4x 2 4 y dy dx 0 32x 2 dx 2304. 14. Since div F ez ez 6 4 0 ez 4 0 y 3e z, we have 6 0 4 6 3ez d V 0 Q 3e z dz dy dx 0 3 e4 y 1 dy dx 0 3 e4 5 dx 18 e 4 5. 15. F x, y, z div F y F S xyi 4 N dS 4y j x xz k div F dV Q 3 0 3 0 3 3 0 3 0 0 0 2 Q y x 4 dV sin 0 2 3 0 sin sin 3 cos 4 2 sin 2 ddd sin2 sin sin2 sin2 cos 4 2 sin ddd 2 sin2 cos 3 sin 4 sin 0 dd 8 0 3 0 2 sin dd 8 0 3 2 2 cos 0 d 16 3 33 16 0 d 144 0 16. div F 2 F N dS Q div F dV Q 2 dV. 4 3 S The surface S is the upper half of a hemisphere of radius 2. Since the volume is 1 2 F S 23 16 3, you have N dS 2 Volume 32 . 3 S ection 15.7 17. Using the Divergence Theorem, we have curl F S Q Divergence Theorem 427 N dS i div curl F dV j y z2 2x 2 6yz k z 2xz 6y i 2z 2z j 4x 4x k 6y i curl F x, y, z 4xy div curl F Therefore, Q x 0. 0. div curl F d V 18. Using the Divergence Theorem, we have curl F S Q N dS i div curl F dV j y yz sin x z k z xyz x sin z 0. xz y sin x i yz xy sin z j yz cos x x cos z k. curl F x, y, z x xy cos z z Now, div curl F x, y, z curl F S y cos x y cos x x sin z 0. Therefore, N dS Q div curl F dV 19. See Theorem 15.12. 20. If div F x, y, z > 0, then source. If div F x, y, z < 0, then sink. If div F x, y, z 0, then incompressible. 21. Using the triple integral to find volume, we need F so that div F M x N y P z 1. y j, or F z k. i 1 fx i 1 fx i 1 x dy dz S S Hence, we could have F For dA dy dz consider F x i, F x i, x f y, z , then N fy j fy2 j fx2 fy j fx2 fz k fz2 fz k fz2 k fy2 and dS 1 fy2 fx2 fx2 fz2 dy dz. fz2 dz d x. fy2 d x dy. For dA dz dx consider F y j, y f x, z , then N and dS 1 For dA dx dy consider F z k, z f x, y , then N and dS 1 Correspondingly, we then have V S F N dS y dz dx S z dx dy. a a a a a 22. v 0 0 x dy dz 0 a a 0 a dy dz 0 a a a 2 dz a 3. a3 Similarly, 0 0 y dz dx 0 0 z dx dy 428 Chapter 15 Vector Analysis 23. Using the Divergence Theorem, we have S curl F N dS Q div curl F dV. Let F x, y, z curl F div curl F Therefore, S Mi P y 2 P xy Nj Pk P x 2 P yx N i z 2 N xz M j z 2 M yz N x 2 N zx M k y 2 M zy 0. curl F N dS Q 0 dV 0. 24. If F x, y, z Therefore, F S a1 i a2 j a 3 k, then div F 0. 25. If F x, y, z F xi N dS yj z k, then div F div F d V Q Q 3. 3 dV 3V. N dS Q div F d V Q 0 dV 0. S 26. If F x, y, z 1 F F S xi yj z k, then div F 1 F Q 3. 1 F Q N dS div F dV 3 dV 3 F Q dV 27. S f D N g dS S fg N dS div f g dV Q Q f div g f g dV Q f 2g f g dV 28. S f DN g gD N f dS S f D N g dS S g D N f dS f g dV Q f Q 2g g 2f g f dV Q f 2g g 2f dV Section 15.8 1. F x, y, z 2y i curl F 2y 3. F x, y, z 2z i i curl F x 2z x z Stokes’s Theorem zi j y xyz 4x2 j j y x yz j k z ez arctan x k k z 2 1 1 x2 j 8xk curl F 4. F x, y, z xyi j yz 2k curl F ezk 2. F x, y, z x2 i i x x2 j y y2 y2 j k z x2 y cos x j j y y cos x yz2 k k z yz 2 2x j x2 k x sin y i i x x sin y z2 i 4x2 arctan x y sin x x cos y k S ection 15.8 5. F x, y, z ex 2 Stokes’s Theorem 1 x2 j k z x2 y2 1 1 1 1 0. y2 y2 k k y2 k 429 y2 i j y ey 2 z2 j x yz k 6. F x, y, z arcsin y i i j y 1 x 1 x 1 1, z y dx C i curl F e x2 k z z2 z2 z2 x y2 curl F x arcsin y e y2 xyz i yz j yz j 2 2ye x xz zx 2 2ze y y2 k y2 2y i 2y i 2e y2 i 2ye x2 k x2 x2 7. In this case, M Line Integral: y F C z, N dr x z, P y x z dx y and C is the circle x 2 x z dy x y dz y2 0, dz x dy C Letting x cos t, y y dx x dy sin t, we have dx 2 sin t d t, dy 2. z2 1. cos t d t and sin 2 t 0 cos 2 t d t x2 y2 C Double Integral: Consider F x, y, z Then N Since z2 1 x2 y 2, z x 2x 2z F F 2x i 2y j 2z k 2 x2 y2 z2 xi yj z k. x z , and z y y z , dS 1 x2 z2 y2 dA z2 1 dA. z Now, since curl F curl F S 2k, we have N dS R 2z 1 dA z 4, z y dx 2 dA R 2 Area of circle of radius 1 2. 8. In this case C is the circle x 2 Line Integral: C y2 0, dz x dy 0. F dr C 2 Let x 2 cos t, y 2 sin t, then dx z x2 2 sin t d t, dy y2 4, N 2 cos t d t, and C y dx k , dS 4y 2 x dy 0 4 dt 1 4x 2 8. 4y 2 d A Double Integral: F x, y, z curl F 2k, therefore F F 2x i 1 2y j 4x 2 2 4 4 x2 2 curl F NdS R 2 2 dA 2 x2 2 dy dx x 2 dx 2x 4 x2 2 2 24 x 2 2 x 2 dx 8. 2 4 2 4 4 arcsin 430 Chapter 15 Vector Analysis z 6 4 9. Line Integral: From the accompanying figure we see that for C1: z C2: x C3 : y Hence, C (0, 0, 6) 0, dz 0, dx 0, dy F dr 0 0 0. 2 C3 C2 2 (0, 3, 0) C1 y x yz d x C y dy y dy C2 0 z dz x 4 (4, 0, 0) y dy C1 3 z dz C3 6 z dz 0 y dy 0 3 y dy 0 z dz 6 z dz 0. Double Integral: curl F Considering F x, y, z N Thus, curl F S xyj 4y xz k 2z 12, then 29 d A. 3x F F 3i 4j 2k and dS 29 N dS R 4 0 4 0 4 4xy ( 3x 0 (12 0 3x) 4 2xz dA 12) 4 4 xy 8xy 0 dx 0 2x 6 3x 2 2y 3 x 2 dy dx 12x dy dx 0. 10. Line Integral: From the figure we see that C1: y C2: z C3: y C4: z Hence, F C z a2 0, z y, x a, z y 2, x 2 0, dy 0, dx a2, dy a, dx dz 0 2y dy 0 a C3 0, dz dz 0, dz C1 C2 a 2y dy. x C4 y dr C z 2 dx 0 dx C1 0 x 2 dy y 2 dz a4 dx C3 C4 a 2y3 dy C2 0 a2 dy 3 2y3 dy 0 a a 2y3 dy a a a4 d x 0 a 2 2y d y y2 z a4 x a2y 0 a5 a3 a3 1 a. Double Integral: Since S is given by N 2yj 1 k and dS 4y 2 2y i 2z j 1 0, we have 4y 2 d A. 2 x k . Therefore, a a a a Furthermore, curl F curl F S N dS R a 4yz a4 0 2x dA 0 0 4yz a 2x dA 0 0 4y3 a3 a3 1 2x dydx a2 . 2ax dx a4x ax2 0 a5 S ection 15.8 \ \ Stokes’s Theorem 431 11. Let A N 0, 0, 0 , B U U V V 1, 1, 1 and C 2 i 2k 22 i 2 0, 2, 0 . Then U k . x AB i j k and V AC 2 j. Thus, Surface S has direction numbers curl F S 1, 0, 1, with equation z 1 2 2 dA R 0 and dS 2 d A. Since curl F 1, b 2 1. 3i j 2 k, we have N dS R dA Area of triangle with h \ \ 12. Let A N 0, 0, 0 , B U U V V x 1, 1, 1 , and C 2i 2 j 22 y and dS i j . 2 0, 0, 2 . Then U AB i j k , and V AC 2 k, and Hence, F x, y, z 2 d A. Since curl F 2x x2 y2 k , we have S curl F N dS R 0 dS 0. 13. F x, y, z z2 i i j x2 j k z y2 z 2y j N dS y 2 k, S: z 4 x2 y 2, 0 ≤ z curl F x z2 x2 2x i y x2 y2 2yi 2z j 2xk G x, y, z G x, y, z 4 k 2 4 4 4 4 x2 curl F S 4xy R 4yz 2x d A 2 2 2 2 x2 x2 4xy 4xy x2 4y 4 16y 0 x2 4x 2y y2 4y 3 2x d y d x 2x d y d x 4x 4 2 x2 dx 14. F x, y, z curl F G x, y, z G x, y, z 4xz i 4x i x2 2x i curl F 4x y2 yj 4 xy k, S: 9 4y j z 9 k x2 y 2, z ≤ 0 2y j N dS 3 9 9 x2 8x 2 R 3 2y 4x x2 4y d A 3 x2 8x 2 x2 32 8xy 8y 2 d y d x S 16x2 9 3 16 9 3 y2 dx 0 15. F x, y, z z2 i i yj j y y 4 4 x x2 x z k, S: z k z xz x2 y2 y2 i 4 4 zj 4 x2 curl F x z2 G x, y, z G x, y, z z y x2 yz x2 y2 y2 j dA k y4 4 x2 x2 y2 dA y2 2 2 4 4 x2 curl F S N dS R y dy dx x2 0 R 432 Chapter 15 x2 i i curl F x x2 G x, y, z G x, y, z z 4 curl F S Vector Analysis z2 j j y z2 4 x x2 N dS R 16. F x, y, z x yz k, S: z k z xyz x2 y2 y2 i 4 xz 4 x2 y2 2z i yzj y x2 zx 2 y2 j k y 2z 4 x2 2 4 4 4 xx 2x x y2 2 2xy x2 x2 4 y 2 dA y2 x2 dA x2 2x y2 dy dx R 2 2 x2 x2y 2 2 y3 3 4 4 x2 x2 dx x2 x2 2 4 3 8 3 4 x 2 x2 4 x2 dx 2x 2 4 2 2 2 4x 4 4x 4 4 1 3 x2 8 82 x4 3 x 2x 2 x2 dx 4 4 3 0 x2 32 81 38 1 8 3 17. F x, y, z ln x 2 i curl F x 1 2 ln x 2 S: z 9 2x 2x 2i y2 y2 i 16 arcsin 4 3 81 32 x4 x2 4 arcsin x 2 2 2 4 2 3 k k z 1 2 x arctan j y j y arctan x y 1 1y x2 y2 y x2 y2 k 2y x2 y2 k 3y over one petal of r 3y 3j N dS R 2 0 2 0 2 2 sin 2 in the first octant. G x, y, z G x, y, z z k 9 curl F S 2y x2 y2 2 sin 2 0 dA 2r sin r dr d r2 2 sin dr d 4 sin cos 0 8 sin 2 cos d 0 8 sin 3 3 2 0 8 3 S ection 15.8 18. F x, y, z yz i i curl F x yz 2 2 j y 3y x 2 3y j k z y2 z2 16 over x 2 y2 16 2y i y 2x j zk x2 y2 k Stokes’s Theorem 433 S: the first octant portion of x 2 G x, y, z G x, y, z curl F S z 16 x 16 x2 x2 i k 2xy 16 x 2 2xy 16 x 2 16 0 x2 N dS R z dA x2 dA 16 x2 y 0 16 R 4 0 4 0 4 2xy 16 x 2 x y2 2 16 16x 64 3 16 x2 dy dx 16 x2 x 16 x 16 dx x2 x2 32 x2 dx x3 3 4 0 0 1 16 3 64 64 3 64 3 19. From Exercise 9, we have N 2x i 1 xz dA R k and dS 4x2 a a 1 4x 2 d A . Since curl F a xyj a5 . 4 x z k , we have curl F S N dS x 3 dy dx 0 0 0 a x 3 dx ax4 4 a 0 20. F x, y, z x yz i i j yj k zk curl F x xyz y y z z xyj xzk S: the first octant portion of z N 2x i 1 curl F S x 2 over x 2 1 4x 2 dA. y2 a 2. We have k and dS 4x2 N dS R a 0 a xz dA R a2 0 x2 x 3 dA x3 dy dx x3 a2 0 x2 dx x2 32 12 2 xa 3 25 a 15 22 a 15 a x2 52 0 434 Chapter 15 i i curl F x 1 Letting N j j y 1 Vector Analysis 2k k z 2 curl F S 21. F x, y, z 22. F x, y, z S: x 0 curl F N dS 0. Letting N 2 zi y 2 yk 1 i x z j y 0 k z y N 0. i j k, we have k, curl F curl F N dS 0 and S 23. See Theorem 15.13. 24. curl F measures the rotational tendency. See page 1031. 25. If curl F F C 0, then Stokes’s Theorem gives dr S curl F N dS 0 The circulation is zero. 26. (a) C fg dr S curl f g f g i x i f g j y j y xfg 2g N d S (Stokes’s Theorem) f g k z k z yfg f y g z f x f x f z g y g z g y i z f 2g fg curl f g fg f x yz f f f y g z j f y g y dr S 2g zy f f f x 2g f z zx g yx g z f z 2 g y f z f y i g x g x g j x j k f x g y f y g k x xz g xy 2 i f x g x Therefore, C k f z g z f g fg dr S curl f g f f f N dS S f g N d S. (b) C ff N d S (using part a) 0. 0 since f —CONTINUED— S ection 15.8 26. —CONTINUED— (c) C Stokes’s Theorem 435 fg gf dr C fg f S dr C gf N dS S dr g f S g g f g N d S (using part a) N dS 0 f S N dS 27. f x, y, z (a) x yz, g x, y, z k z, S: z 4 x2 y2 g x, y, z f x, y, z g x, y, z rt 2 cos t i x yz k 2 sin t j dr xyk 0 k, 0 ≤ t ≤ 2 0 f x, y, z g x, y, z C (b) f x, y, z g x, y, z f g yz i k i yz 0 x x2 1 xz j j xz 0 y2 4 i k xy 1 4 x x2 y2 xz i y x2 2 yz j N dS 4 y2 j k y x2 S 2 4 N dS y2 dA x 2z 4 x2 4 y2 2 x2 y2 dA 2 x2 dA f x, y, z S g x, y, z y 2z 4 x2 y2 4 y2 S 2 0 2 0 2 0 2 x2 y2 dA 4 x2 y2 2r 2 cos 2 4 sin 2 r2 2 0 r d dr dr 0 2r 3 1 sin 2 4 r2 2 28. Let C 1 2 since C and ai C C bj r c k, then dr 1 2 curl C S r N dS 1 2 2C S N dS S C N dS r i a x j b y k c z bz cy i az cx j ay bx k i curl C r bz x cy cx j y az ay k z bx 2 ai bj ck 2C. 436 Chapter 15 Vector Analysis z 29. Let S be the upper portion of the ellipsoid x2 4y 2 z2 4, z ≥ 0 S: z = 4 − x 2 − 4y 2 2 Let C : r t If F 0 S 2 cos t, sin t, 0 , 0 ≤ t ≤ 2 , be the boundary of S. 1 M, N, P exists, then curl F F C N dS (by (i)) x 2 1 y C dr dr sin t 2 cos t , ,0 4 4 (Stokes’s Theorem) (by (iii)) 2 sin t, cos t, 0 dt G C 2 0 1 4 2 2 sin2 t 0 2 cos2 t dt Hence, there is no such F. Review Exercises for Chapter 15 1. F x, y, z z 3 2 5 4 3 2 x −1 −2 −3 −4 −5 2 4 xi j 2k 2. F x, y i y 2y j 2 3 x 4 y 3. f x, y, z F x, y, z 8x2 16x xy yi z2 xj 2z k 4. f x, y, z F x, y, z x 2eyz 2xeyz i xeyz 2i x 2ze yz j xz j x 2 ye yz k xyk 5. Since M y 1 y2 N x, F is not conservative. 1 x, partial integration 1 x2 N x, F is conservative. From M Ux y x 2 and N Uy 6. Since M y yields U yx h y and U yx g x which suggests that U x, y yx C. N x, F is conservative. From M U 7. Since M y 12xy partial integration yields U 3x 2 y 2 x3 h y and U 3x 2 y 2 gx x3, and U x, y 3x 2 y 2 x3 y3 7y C. x 6xy 2 y3 7y 3x 2 and N U y 6x 2 y 3y 2 7, g x which suggests h y y3 7y, 6y 2 sin 2x N x, F is conservative. From M Ux 2y3 sin 2x and N 8. Since M y g x which suggests that h y y3, g x we obtain U y3 cos 2x h y and U y3 1 cos 2x U x, y y3 1 cos 2x C. U y 3y 2 1 C, and cos 2x , R eview Exercises for Chapter 15 9. Since M y 4x N , x 10. Since M y M z N z 4x 2z 6y N , x P , x P y 437 M P 1 . z x F is not conservative. F is not conservative. 11. Since M y 1 y 2z N , x M z 1 yz2 P , x N z x y 2z2 P , y F is conservative. From M we obtain U x yz f y, z , U x yz g x, z , U x yz h x, y ⇒ f x, y, z x yz K. U x 1 ,N yz U y x ,P y 2z U z x yz2 12. Since M y sin z N , x M z y cos z P , x F is not conservative. 13. Since F (a) div F (b) curl F x2i 2x P y xy 2 j 2 xy 2 xz j cos y y 2j 2y z2 k: 2z N i z z x 2 k; x2 y 2k y cos x i x cos y cos x y 1 3j k 18. Since F (a) div F (b) curl F x2 2x 0 yi x sin2 y j: 5 2z j sin x xy sin y z sin y 3x k: cos x k xz i yz j x sin y j x yz k: P x M j z N x M k y 0i 0j 0k 0 14. Since F (a) div F (b) curl F 15. Since F (a) div F (b) curl F 16. Since F (a) div F (b) curl F 17. Since F y sin x xz i 3x 3 2i yz j yi 1 arcsin x i x y 2 j y z2 k: 1 (a) div F 2xy 2yz 1 x2 (b) curl F z2 i y2k 2 sin y cos y 438 Chapter 15 Vector Analysis z i x z x2 1 i y z j y z y2 1 j x 19. Since F (a) div F ln x 2 2x x2 2x x2 y2 i y2 2y y2 2y k y2 t, x2 1 ln x 2 2y y2 y2 j 1 z k: 20. Since F (a) div F (b) curl F z 2 k: 2z z2 1 x2 1 y2 (b) curl F 2x x2 t, y 21. (a) Let x x2 C 1 ≤ t ≤ 2, then ds 2 2 dt. 22 t3 3 2 y 2 ds 1 2t 2 2 dt 62 1 (b) Let x x2 C 4 cos t, y y 2 ds 4 sin t, 0 ≤ t ≤ 2 , then ds 2 4 dt. 16 4 dt 0 128 22. (a) Let x 5t, y xy ds 4t, 0 ≤ t ≤ 1, then ds 1 41 dt. 4 3 2 y 20t 2 0 41 dt C 20 41 3 dt 2 5 dt dt 2 (b) C1: x C2: x C3: x t, y 4 0, y 0, 0 ≤ t ≤ 4, ds 4t, y 2 xy ds C 0 (0, 2) y = − 1x + 2 2 C2 (4, 0) x 3 4 2t, 0 ≤ t ≤ 1, ds t, 0 ≤ t ≤ 2, ds 4 1 C3 C1 Therefore, 0 dt 0 8t t2 2 t3 3 1 0 8t 2 2 5 dt 0 0 dt 16 5 85 . 3 dx dt t cos t, t cos t 2 23. x cos t x2 C t sin t, y y 2 ds sin t 2 t cos t, 0 ≤ t ≤ 2 , cos t t sin t 2 2 dy dt t sin t 2 sin t t 2 cos2 t t 2 sin2 t dt 0 t3 t dt 0 2 2 1 2 24. x t sin t, y 2 1 t 0 2 cos t, 0 ≤ t ≤ 2 , sin t 1 cos t 2 dx dt 1 cos t, dy dt 2 sin t t sin t 2 1 3 2 2 cos t dt 2 2 x ds C sin t 2 dt 0 2 0 2 t1 cos t sin t 1 cos t dt 2 cos t 32 0 2 0 t1 cos t dt 2 0 t1 cos t dt 8 R eview Exercises for Chapter 15 25. (a) Let x 2t, y 2x C 439 3t, 0 ≤ t ≤ 1. 1 1 y dx x 3y dy 0 7t 2 3 sin t dt, dy 2 7t 3 dt 0 35t dt 35 2 (b) x 3 cos t, y 2x C 3 sin t, dx x 3y dy 3 cos t dt, 0 ≤ t ≤ 2 9 sin t cos t dt 18 y dx 9 0 26. x cos t 2x C t sin t, y y dx x sin t t sin t, 0 ≤ t ≤ 2 2 , dx t cos t dt, dy 6t 2 cos t cos2 t t t cos t 1 sin t dt 3 dt 1.01 3y dy 0 sin t cos t 5t 2 sin2 t 2t 27. C 2x xt yt 2x C y ds, r t 3a 3a a cos3 t i a sin3 t j, 0 ≤ t ≤ 2 cos2 t sin t sin2 t cos t 2 y ds 0 2a cos3 t a sin3 t xt 2 y t 2 dt 9 a2 5 28. r t xt x2 C ti t2j t 3 2 k, 0 ≤ t ≤ 4 2t, z t 4 1, y t y2 31 t 2 t2 2 z2 ds 0 t4 t3 1 4t 2 9 t dt 4 2080.59 29. f x, y C: y rt rt rt 5 sin x y 3x from 0, 0 to 2, 6 ti i 3t j, 0 ≤ t ≤ 2 3j 10 Lateral surface area: 2 2 f x, y ds C 0 5 sin t 3t 10 dt 10 0 5 sin 4t dt 10 41 4 2t i 3t 2 j dt cos 8 32.528 30. f x, y C: y rt rt rt 12 x y 31. d r F F C x 2 from 0, 0 to 2, 4 ti i t 2 j, 0 ≤ t ≤ 2 2t j 1 4t 2 t5 i dr t 4 j, 0 ≤ t ≤ 1 1 5t 6 d t 0 5 7 Lateral surface area: 2 f x, y ds C 0 12 t t2 1 4t 2 dt 41.532 440 32. dr F Chapter 15 4 sin t i 4 cos t 2 Vector Analysis 3 cos t j dt 4 cos t 3 sin t j, 0 ≤ t ≤ 2 12t 7 sin2 t 2 2 3 sin t i 12 0 F C dr 7 sin t cos t dt 24 0 33. d r F F C 2 sin t i 2 cos t i 2 2 cos t j 2 sin t j k dt 34. x dr F F C 2 t, y i 4 dr 0 2 j t, z 4t t 2, 0 ≤ t ≤ 2 t k, 0 ≤ t ≤ 2 2 2t k dt 4t t 2 4t t2 i 2 dt t2 2 4t 2 dr 0 t dt 2 2t 2 t2 2 2 tj 0k t 2t 0 35. Let x F F C t, y t dr 2 t, z 2t 2 i 2 2t 2, 2t2 tj 2 ≤ t ≤ 2, d r 2t k 64 3 i j 4 t k dt. 4t2 dt 4t3 2 3 2 36. Let x dr F F C 2 sin t, y 2 cos t i 2 cos t, z 2 sin t j 2 sin t k 8 sin t 4 sin2 t, 0 ≤ t ≤ 8 sin t cos t k dt . 0i dr 4j 16 sin2 t cos t dt 8 cos t 0 16 3 sin t 3 16 0 37. For y For y xy dx C x 2, r1 t 2x, r2 t x2 ti 2 y 2 dy t 2 j, 0 ≤ t ≤ 2 ti 4 xy dx C1 y y = 2x 4 3 2t j, 0 ≤ t ≤ 2 x2 32 4 3 y 2 dy C2 (2, 4) C2 y = x2 C1 x 1 2 3 4 xy dx x2 y 2 dy 2 100 3 1 38. C F rt F C dr 2 cos t dr 4 C 2x y dx 2y 2 sin t x dy 2 t cos t j, 0 ≤ t ≤ 2 t sin t i 2 4 39. F xi y j is conservative. 12 x 2 23 y 3 4, 8 2 0, 0 Work 1 16 2 23 8 3 2 8 3 3 42 R eview Exercises for Chapter 15 2000 5280 t k, 0 ≤ t ≤ 2 2 25 tk 33 1, 4, 3 441 40. r t 10 sin t i 10 sin t i 10 cos t j 10 cos t j 41. C 2xyz dx x 2z dy x 2 y dz x 2 yz 0, 0, 0 12 F dr F C 20k 10 cos t i 2 10 sin t j 500 dt 33 1 dz z 25 k 33 250 mi 33 ton dr 0 4, 4, 4) 42. C y dx x dy xy ln z 0, 0, 1 16 ln 4 1 43. (a) C y 2 dx 2xy dy 0 1 1 3 t2 0 1 t 2 3 1 21 3t 1 2 3t 2 4t t dt 1 dt 2t 14t 7t 2 9t 2 0 5 dt 1 3t 3 4 5t 0 15 t 1 2t dt (c) F x, y Hence, y 2i 2 xy j f where f x, y xy 2. (b) C y 2 dx 2xy dy 1 4 t1 t 1 4 2t t dt F 15 C dr 42 2 11 2 15 t2 1 44. x a 1 2 sin ,y a1 y dx. cos ,0 ≤ ≤2 y (a) A x dy C Since these equations orient the curve backwards, we will use A 1 2 1 2 a2 2 a2 2 2 C1 C2 2a 2π a y dx x dy 1 2 x a2 1 0 2 cos 1 cos a2 sin sin d 0 0 0d 1 0 2 2 cos cos2 sin a2 6 2 sin2 d 2 0 2 cos sin d 3 a2. (b) By symmetry, x y 1 2A a. From Section 15.4, y 2 dx 1 2A 5 a 6 2 a3 1 0 cos 2 1 cos d C 1 a3 5 2 3 a2 442 Chapter 15 Vector Analysis 2 2 2 2 2 45. C y dx 2x dy 0 0 2 1 dy dx 0 2 dx 4 46. C xy dx x2 y 2 dy 0 2 0 2x 2x dx 0 x dy dx 4 a a2 a2 x2 47. C xy 2 dx x 2y dy R 2xy 2xy dA 0 48. C x2 y 2 dx 2xy dy a a x2 4y dy dx 0 dx a 0 1 x 1 49. C xy dx x 2 dy 0 x2 x dy dx 0 x2 x3 dx 1 12 1 1 (1 x2 33 2 50. C y 2 dx x 4 3 dy 1 1 1 1 1 x2 3 3 2 41 x 3 1 1 3 2y dy dx 332 4 13 x y y2 3 81 x 3 82 x 7 3 x2 dx x2 332 1 1 x2 x2 332 dx 16 1 35 1 3 352 x2 352 1 0 u k 6 z 51. r u, v 0≤u≤ sec u cos v i 3 , 1 2 tan u sin v j z 6 2u k 52. r u, v e u4 cos v i e u4 sin v j 0≤v≤2 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 2 −4 2 4 x −2 2 4 y x 2 2 y 53. (a) 3 −4 4 x −2 −3 z (b) 3 −4 −4 −3 4 y x 4 3 2 −1 −2 −3 z −4 2 3 4 y (c) 3 2 −4 −3 4 x 3 −2 −3 z (d) 3 −2 2 −3 −4 −4 −3 3 4 y x 4 3 −2 −3 1 z −4 −2 2 3 4 y The space curve is a circle: r u, —CONTINUED— 4 32 cos u i 2 32 sin u j 2 2 k 2 R eview Exercises for Chapter 15 53. —CONTINUED— (e) ru rv ru rv 3 cos v sin u i 3 sin v cos u i 3 cos v cos u j 3 sin v sin u j cos v k 443 i 3 cos v sin u 3 sin v cos u 3 cos2 v cos u i 3 cos2 v cos u i j k 3 cos v cos u 0 3 sin v sin u cos v 3 cos2 v sin u j 3 cos2 v sin u j 9 cos4 v sin2 u 9 cos v sin v sin2 u 9 cos v sin v k 81 cos2 v sin2 v 9 cos v sin v cos2 u k ru rv 9 cos4 v cos2 u 9 cos4 v 81 cos2 v sin2 v Using a Symbolic integration utility, 2 4 2 ru 0 rv du dv 14.44. (f) Similarly, 4 0 0 2 ru rv dv du 4.27. 54. S: r u, v ru u, v rv u, v ru ru rv rv z dS S u i i i 1 1 j j vi u vj sin v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ cos v k cos v i cos v j 2k j k 10 1 cos v 2 cos2 v 2 4 4 du dv 2 6 2 ln 6 6 2 2 z 2 −3 3 x 3 −2 y −3 sin v 2 cos2 v 0 0 55. S: r u, v ru u, v rv u, v ru ru rv rv x S u cos v i cos v i u sin v j sin v j 3 u 2u k 12 u k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 u sin v i i cos v u sin v u y dS 0 2 0 u cos v j j k sin v 3 2u u cos v 0 3 2 2u 3 u cos v i 2u 3 u sin v j uk 2u 2 1 u cos v u sin v u 2u 3 2 2 1 du dv 0 2 cos v 0 sin v u2 2u 3 2 1 dv du 0 444 Chapter 15 aa 0 ⇒ x2 z k x2 x2 Vector Analysis y2 , 0 ≤ z ≤ a2 y2 a2 y2 a a y z 56. (a) z z a2 a x2 y2 a 2 (b) S: g x, y x, y m S e x, y, z dS k x2 R x y2 y2 1 2 a 1 1 x2 gx2 gy2 d A a2y2 dA x y2 2 k R x2 a2 R a2x2 x y2 2 k k a2 k a2 2 k a2 3 57. F x, y, z x2 i y2 dA 1 0 2 0 r2 dr d a3 d 3 1 0 1 a3 xy j zk 3y 4z 12 Q: solid region bounded by the coordinates planes and the plane 2x Surface Integral: There are four surfaces for this solid. z 0 N k, F N z, S1 0 dS 0 y 0, N j, F N xy, S2 0 dS 0 x 0, N i, 2i F N x 2, S3 0 dS 1 4 0 9 dA 16 29 dA 4 2x 3y 4z 12, N 1 4 1 4 1 4 1 6 1 6 3j 4k , dS 29 3xy 2x 2 4z dA 3xy 12 1 F S4 N dS 2x 2 R 6 0 6 4 0 (2x 3) 2x 2 3y dy dx 12 12 3 2x 2x 12 3 2x 3 12 2x 2 3 2 2x 2 0 6 12 3 x2 x3 3 2x 3x 12 2x 2 3 24x 36 dx 6 dx x3 0 x4 4 12x 2 36x 0 66 —CONTINUED— R eview Exercises for Chapter 15 57. —CONTINUED— Divergence Theorem: Since div F 6 (12 0 (12 0 6 2x) 3 2x) 3 445 2x (12 0 2x x 3y) 4 1 3x 3x 1, Divergence Theorem yields div F dV 0 Q 6 0 1 dz dy dx 3y 3x 1 4 1 4 1 4 58. F x, y, z xi yj 1 12 2x 4 32 y 2 2x dy dx 2x 3 (12 0 3x 0 6 1 12y 2xy dx 12 3 1 3x 4 64 2x 3 12 2x 2 3 35x3 3 2 3x 0 6 0 1 4 12 2x dx 6 23 3x 3 35x 2 96x 36 dx 48x 2 36x 0 66. zk 3y 4z 12 (0, 0, 3) z Q: solid region bounded by the coordinate planes and the plane 2x Surface Integral: There are four surfaces for this solid. z 0 N k, F N z, S1 0 dS 0 (0, 4, 0) y x y 0, N j, F N y, S2 0 dS 0 (6, 0, 0) x 2x 3y 0, 4z N i, 12, N 1 4 1 4 2i F N x, S3 0 dS 1 1 4 0 9 dA 16 29 dA 4 3j 4k , dS 29 3y 4z dy dx 6 N S4 F dS 2x R 6 0 (12 0 2x) 3 12 dy dx 3 0 4 2x dx 3 3 4x x2 3 6 36 0 Triple Integral: Since div F div F dV Q Q 3, the Divergence Theorem yields. 3 Volume of solid 3 1 Area of base Height 3 1 643 2 36. 3 dV 59. F x, y, z cos y 2 y cos x i sin x x sin y j x yz k S: portion of z y over the square in the xy-plane with vertices 0, 0 , a, 0 , a, a , 0, a z 1 Line Integral: Using the line integral we have: C1: y C2: x C3: y C4: x 0, dy 0 C3 0, dx a, dy a, dx 0, 0, 0, z z z y 2, a2, y 2, dz dz dz 2y dy 0 2y dy x C4 C1 a C2 a y —CONTINUED— 446 Chapter 15 Vector Analysis 59. —CONTINUED— F C dr C cos y y cos x dx sin x x sin y dy xyz dz dx C1 a C2 0 0 C3 cos a a cos x dx C4 a sin a a sin y dy a ay3 2y dy dx 0 a cos a x cos a a cos x dx 0 0 sin a a a sin y dy 0 2ay 4 dy a 0 a a a sin x a y sin a a sin a z 1 a cos y 0 2a 2a6 5 y5 5 a cos a a sin a a cos a a 2a6 5 Double Integral: Considering f x, y, z N Hence, a a y 2, we have: xz i y z j. f f 2y j k , dS 1 4y 2 4y 2 dA, and curl F a a a curl F S N dS 0 0 2y 2z dy dx 0 0 2y 4 dy dx 0 2a5 dx 5 2a6 . 5 60. F x, y, z x zi y zj x2 k y 2z 12 (0, 0, 6) z S: first octant portion of the plane 3x Line Integral: C1: y C2: x C3: z F C 0, 0, 0, x C dy dx dz 0, 0, 0, y 3x 2 5 x 2 z z y 12 2 12 2 12 z dy x2 6 dx 3 2 3x y , dz dz dy 3 dx 2 1 dy 2 3 dx x (4, 0, 0) (0, 12, 0) y , 3x, x 2 dz dx dr z dx 12 32 x 2 x C1 0 4 y C2 12 2 y 4 dy C3 x 36 dx 12 8 3x 3 dx 12 0 3 y 2 z k 6 dy 0 10x Double Integral: G x, y, z G x, y, z curl F i 12 3x 2 3 i 2 y 1 j 2 2x N dS 1j 4 12 0 3x 4 curl F S x 0 1 dy dx 0 3x 2 15x 12 dx 8 61. If curl F xi yj zk, then div curl F contradicting Theorem 15.3. 1 1 1 3, Problem Solving for Chapter 15 447 Problem Solving for Chapter 15 1. (a) T N dS Flux S 25 x2 xi 1 1 x2 y2 1 z2 32 xi yi zk x2 k dA N dS x2 z2 3 y2 1 1 1 y2 3 y2 32 32 kT 25k R 12 x2 1 y2 x2 2 1 2 x2 1 12 12 x2 12 z y2 x2 z2 y2 32 dA dy dx 25k 12 0 12 1 x2 z2 3 1 12 x2 1 x2 z2 3 2 1 x2 12 25k 12 0 1 x2 1 dy dx dx 25k 0 1 2 2 dy 12 1 x2 12 25k (b) r u, v ru ru T T Flux 0 25k 2 6 cos u, v, sin u sin u, 0, cos u , rv 0, 1, 0 rv x2 ru 1 cos u, 0, 25 y2 rv 2 3 3 sin u xi yj zk v2 sin2 u 2 6 z y 1 x2 y x2 25 13 2 z2 v2 v2 32 cos u i 25 1 vj sin u k 25 13 32 2 cos2 u 25k v2 32 25k 1 z x z y du dv 2. (a) z dS T 1 x2 z x 2 y2, 1 2 x x2 y2 1 zk , 1 y2 y2 1 dA xi k yi 25 x2 y2 z i x z x xi yj kT S 2 z2 32 25 x i yj zk N z j y z y 1 2 1 1 x x2 y2 i 1 y x2 y2 j k 1 x2 y2 x2 N dS y2 k k R xi 25 x i 2 yj yj 1 0 zk zk 1 1 r2 xi r dr d yj zk 50 k 1 x2 dA Flux k R 1 y2 25 1 x2 dA y2 25k 0 —CONTINUED— 448 Chapter 15 Vector Analysis 2. —CONTINUED— (b) r u, v ru rv ru ru Flux rv rv 25k 0 0 sin u cos v, sin u sin v, cos u cos u cos v, cos u sin v, sin u sin u sin v, sin u cos v, 0 sin2 u cos v, sin2 u sin v, sin u cos u sin2 v sin u 2 2 sin u cos u cos2 v sin u du dv 50 k 3. r t rt Ix C 3 cos t, 3 sin t, 2t 3 sin t, 3 cos t, 2 , r t 2 13 4t2 4t2 13 dt 13 dt 1 3 1 3 13 13 32 32 2 y2 x2 C z2 z2 ds 0 2 9 sin2 t 9 cos2 t 0 2 27 27 Iy ds 2 Iz C x2 y2 ds 0 9 cos2 t 9 sin2 t 13 dt 18 13 4. r t rt ds Iy C t2 2 2t 3 , t, 2 3 t, 1, 1 1 x2 y2 C 2 2t1 t t z2 z2 y2 2 ,rt 1 1 t 1 1 dt ds 0 1 t4 4 t2 83 t dt 9 83 t dt 9 t 2 dt 49 180 5 9 23 60 Ix Iz C ds 0 1 x2 ds 0 t4 4 5. 1 x dy 2C y dx 1 2 2 a 0 2 sin sin a sin sin2 2 cos d 1 a1 2 cos 2d cos a1 d cos d 12 a 2 12 a 2 cos2 0 2 sin 0 3 a2 Hence, the area is 3 a2. 1 x dy 2C 2 6. y dx 2 0 1 sin 2t cos t 2 sin t cos 2t dt 2 2 3 Hence, the area is 4 3. P roblem Solving for Chapter 15 7. (a) r t rt W (b) r t rt W F t j, 0 ≤ t ≤ 1 j 1 1 449 F C dr 0 ti j j dt 0 dt 1 t 1 dr t2 i 2t i 1 t j, 0 ≤ t ≤ 1 j 2t 0 1 t2 i 2t 2t 4 t2 2t t t2 t2 2 1j 2t3 13 15 1 t2 2t i 1 dt j dt 1 0 1 t4 1 dt t4 0 (c) r t rt F dr ct c1 ct c2t 4 t2 i 2t i t2 t j, 0 ≤ t ≤ 1 j t c1 c2t 1 c 6 5 2 2t 2 ct 2 1 c2 t ct t2 1 2 11 2c2t 2 12 c 30 W dW dc d 2W dc 2 8. F x, y f x, y Work F C dr 1 6 1 c 15 0⇒c c 1 >0 15 3x2 y 2 i x3y2 f 2, 4 5 minimum. 2 9. v r a1, a2, a3 a2z 1 127 curl v r a 3 y, x, y, z a1z a 3 x, a 1 y 2v a2 x 2 x 3 y j is conservative. potential function. f 1, 1 8 16 2 a 1, 2 a 2, 2 a 3 By Stokes’s Theorem, v C r dr S curl v 2v S r N dS N dS. 10. Area rt rt F F dr 2 ab a cos t i a sin t i b sin t j, 0 ≤ t ≤ 2 b cos t j 1 a cos t j 2 1 ab cos2 t dt 2 ab 1 ab 2 1 b sin t i 2 1 ab sin2 t 2 F 0 W dr 1 ab 2 2 Same as area. 450 Chapter 15 Vector Analysis m y2 y2 2y x2 x2 x2 y2 x2 12y 2 72 11. F x, y M M y M x, y i 3mxy y2 5 N x, y j x2 52 3xy i 2y 2 x2 j x2 2 3mxy x 2 y2 72 72 52 3mxy 3mx x 2 52 x 2 y2 x2 y2 y2 2x 5 y2 52 3mx 5y 2 3 mx x 2 4 y 2 x2 y2 7 2 52 N N x m 2y 2 x 2 x2 y2 5 2 m 2y 2 mx x2 mx x2 x2 y2 y2 m 2y 2 52 x 2 72 x2 x2 y2 y2 2 52 2 mx 2y 2 3x 2 72 3mx x2 x2 y2 4y2 72 Therefore, N x M and F is conservative. y ...
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