EVEN14 - C H A P T E R 14 Vector Analysis Section 14.1...

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Unformatted text preview: C H A P T E R 14 Vector Analysis Section 14.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 419 Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 423 Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 427 Section 14.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 431 Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 436 Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 439 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 CHAPTER Vector Analysis Section 14.1 14 Vector Fields Solutions to Even-Numbered Exercises 2. All vectors are parallel to x-axis. Matches (d) 4. Vectors are in rotational pattern. Matches (e) 6. Vectors along x-axis have no x-component. Matches (f) 8. F x, y F 2 2i 10. F x, y F xi x2 y 5 4 3 2 1 yj y2 y 1 x 1 2 −2 −1 −5 − 4 −2 −1 −2 −3 −4 −5 x 12 45 −2 12. F x, y F x xi c y 3 2 1 14. F x, y F y 10 8 6 x 1 2 3 x2 1 y2 i x2 y2 j 2 4 −3 −2 −1 −2 x 2 16. F x, y, z F x2 y2 x 2 xi y z2 z yj 2 zk z 2 18. F x, y 6 4 2 −6 −4 −2 2y y 3x i 2y 3x j c c2 x 2 4 6 2 −2 −2 −6 2 x −2 2 y 407 408 Chapter 14 xi z Vector Analysis yj zk 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 20. F x, y, z 2 1 1 2 x 1 2 y 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 26. g x, y, z gx x, y, z gy x, y, z x arcsin yz arcsin yz xz 1 xy 1 y2z2 xz 1 y2z2 j xy 1 y2z2 k y2z2 1x x y z i y gz x, y, z 1 z xz j y2 y z2 1 x x k y G x, y, z arcsin yz i 28. F x, y 1 yi x2 xj y i x2 1 j x 30. F x, y 1 yi xy xj 1 i x 1 j y M y x2 and N derivatives for all x N x 1 x2 1 x have continuous first partial 0. 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 M ⇒ F is conservative. y 32. M N x x x2 x2 y2 ,N xy y2 y x2 32 y2 M ⇒ Conservative y 34. M N x y 1 1 x2y2 ,N 1 32 x x2y2 M y 1 1 x2y2 32 1 x2y2 ⇒ Not conservative x2 j y2 36. F x, y 1 yi y2 1 i y 2xj 2x j y2 38. F x, y y x 3 x2 y2 2x3y 3 x2 y2 i 6x2 y 6x2y 2x3yj 40. F x, y 2y yx x x2 y2 2y i x 2 x 1 yy x 2x y2 1 y2 2 y2 2x y2 Conservative fx x, y fy x, y f x, y 3 x2 y2 2x3y x3y2 K Not conservative Not conservative Section 14.1 2x x2 2x y x2 x x2 y2 2y y2 2 2 Vector Fields 409 42. F x, y y2 2 i 2y x2 8 xy x2 y2 8 xy x2 y2 3 y2 2 j 44. F x, y, z x2z i i j 2 xz j k y z k, 2, 1, 3 curl F x x2z z y z 2xz yz 2x i 2x i 0 x2 j 7i 4j x2 j 2z k 6k 2z 0k 3 Conservative fx x, y fy x, y f x, y 2x x2 2y x2 1 x2 y2 y2 2 z y2 2 curl F 2, 1, 3 K 46. F x, y, z e i x yz i j y e 6i 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 x yz curl F 3, 2, 0 48. F x, y, z yz y i z i j xz x z j k z xy xy x y k curl F y x yz z x2 x x2 x x2 i y 1 y 2 y xz 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 z2 i z 2 1 1 z z2 y x z 2 50. F x, y, z y2 j j y y2 k k y zi z y2 z x2 xj x y2 z2 yk curl F x2 x y2 z2 x2 z2 x2 z2 52. F x, y, z ez y i i j xj k k 54. F x, y, z y 2z 3 i i j 2 xyz 3 j k 3 xy2z 2 k curl F x yz z ye xez ez xe z i yezj 0 curl F y x 2z 3 y z 2 xyz 3 3xy 2z2 0 Not conservative Conservative f x, y, z x y 2z3 K 410 Chapter 14 x x2 i curl F x2 Conservative fx x, y, z fy x, y, z fz x, y, z f x, y, z x x y2 Vector Analysis i y x2 j y y y2 x2 y2 y2 k z 1 0 j k xex i x xe x ex x xex ex 1 yey j y ye y ey y yey ey 1 56. F x, y, z 58. F x, y div F x, y x x2 y x2 1 x x2 1 ln x 2 2 y2 dx y2 dy y2 z h x, z K3 K K2 g y, z K1 y2 y2 f x, y, z y x2 1 ln x 2 2 y2 f x, y, z f x, y, z dz 1 ln x 2 2 ln x 2 x p x, y y2 z 60. F x, y, z div F x, y, z y2 i y2 2 xz j y xy j y yz k ln y 2 xy z2 k z ln y 2 z2 2x x2 64. y2 x 2z y2 z2 ln x yz i j k 111 x y z 11 11 1 32 6 ln x 2 x 2z i 2 xz 11 62. F x, y, z div F x, y, z div F 2, 1, 3 F x, y, z div F x, y, z div F 3, 2, 1 66. See the definition of Conservative Vector Field on page 1011. To test for a conservative vector field, see Theorem 14.1 and 14.2. 70. F x, y, z G x, y, z F G xi xi ij x0 x2 y i curl F G x yz x xx 2 68. See the definition on page 1016. zk yj k z z2 z2 k yz i j y xz2 2 xz 2z x 2z x2 i 1i xz2 k z xy y zz yj 2x 1k z2 2 xz zk x2z j xyk Section 14.1 72. F x, y, z x2z i i curl F x x2z j y 2xz i curl curl F z 76. F x, y, z x2z i i curl F x 2 xz 2 j x 2x 2 xz j k z 2x i x2 j 2z k 2 xz j k z yz j y x2 yz k z 2x i x2 j 2z k F k z 2z j 2 xk G G yz k 74. F x, y, z G x, y, z xi x2 i i x x2 j 0 y 0 zk yj k z z2 z2k yzi Vector Fields 411 xz2 x 2z j xyk div F y z 2xz yz 2 0 div 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 80. Let F F Mi G Nj j y f y f k z k z f z Ri PS i Sj T k. MT PR j MS NR k 2f 2f yz zy i 2f 2f xz zx j 2f 2f xy yx k 0 P k and G NT 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 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 412 Chapter 14 Mi curl F div curl F Nj Vector Analysis Pk. P y x 2P 84. Let F N i z P y N z 2N P x y 2P M j z P x 2M N x z 2N M k y N x 2M M z yz zx M y 0 (since the mixed partials are equal) xy xz xi yx yj zy In Exercises 86 and 88, F x, y, z 1 f 1 f 1 f x2 x2 x2 1 y2 x x2 y2 z2 1 y2 z2 z2 z2 z k and f x, y, z F x, y, z x2 y2 z 2. 86. 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 88. w w x w y w z 2w x2 x y2 y y2 z y2 z2 32 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 y2 2w 32 z2 2w 32 x2 y2 z2 0 Therefore w 1 is harmonic. f Section 14.2 x2 16 cos2 t y2 9 sin2 t cos2 t sin2 t x y rt Line Integrals ti 5i 14 4 5 t j, 2. 1 1 x2 16 y2 9 4 cos t 3 sin t 4 cos t i 3 sin t j 4. r t 0≤t≤5 9 t j, 5 ≤ t ≤ 9 t i, 9 ≤ t ≤ 14 0≤t≤2 ti 4 8 ti 4 xy d s C 0 6. r t t 2 j, ti t j, 2 2 0≤t≤2 4 j, 2 ≤ t ≤ 4 4≤t≤8 t j, 0 ≤ t ≤ 2; r t i 42 0 8. r t j 2 4t 2 t 1 1 dt 2t t 2 dt 4 2 t2 t3 3 2 4 24 0 8 3 16 2 3 Section 14.2 10. r t 12 t i 8 xyz d s C 0 Line Integrals 413 5t j 2 3 k, 0 ≤ t ≤ 2; r t 122 52 12 i 02 d t 5j 2 8 12t 5t 3 18,720 t 2 d t 0 18,720 t3 3 2 49,920 0 12. r t x2 C 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 14. r t x2 C 2 cos t i y2 ds 2 sin t j, 0 ≤ t ≤ 2 y 2 2 4 cos2 t 0 2 4 sin2 t 2 sin t 2 2 cos t 2 d t 1 8 dt 0 4 1 2 x 16. r t x C ti 3 t j, 0 ≤ t ≤ 3 3 9 y (3, 9) 4 y ds 0 t 4 3t 1 2 9 dt 6 t2 10 2 10 27 6 ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ 2 8 33 t 3 144 3 0 3 57 10 2 −3 x 3 6 18. r t t i, 2i 6 8 x 0 t 2 j, 2 t i 2 j, 4 t j, 6 2 t t t t 2 4 6 8 y 2 C3 (2, 2) C4 1 C2 4 y ds 0 4 t dt C1 C1 x 1 2 x C2 4 y ds 2 6 2 4t 2 ds 4 16 2 3 2 82 x C3 4 y ds 4 8 6 48 6 t 42 t ds 16 2 3 ds x C4 4 y ds 16 2 3 2 82 16 2 3 8 56 3 2 x C 4 y ds 2 4 414 20. Chapter 14 x, y, z rt rt rt Mass C Vector Analysis 22. F x, y 3 sin t j 3 cos t j 2 4 z 3 cos t i 3 sin t i 3 sin t 2 t k, 0 ≤ t ≤ 4 2k 2 xyi yj 4 sin t j, 0 ≤ t ≤ 4 sin t j 2 C: r t Ft 4 cos t i 16 sin t cos t i 4 sin t i F dr 0 3 cos t 2 2 13 16 2 rt 13 C 4 cos t j 2 x, y, z d s 0 2t 13 d t 64 sin2 t cos t 64 3 sin t 3 8 sin2 t 0 16 sin t cos t dt 2 40 3 24. 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 26. F x, y, z C: r t Ft rt 4t dt t2 2 2 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 2 cos t j 4 cos2 t j 2 sin t j 4 4 t2j t j 4 t2 2 4 sin2 t i 2 cos t i F dr C 0 3t 2 0 2 C 8 sin2 t cos t 83 sin t 3 8 3 6 15 t dt 4 8 cos3 t 3 8 3 24 0 24 16 3 28. 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 e t k, 0 ≤ t ≤ 2 et k e2t et k dt 2 1 2t 2 e2t 2t e 2 t dt 6.91 30. F x, y C: x rt rt Ft F r x2 i cos3 t, y cos3 t i 3 xyj sin3 t from 1, 0 to 0, 1 sin3 t j, 0 ≤ t ≤ t sin t i 3 sin2 t 2 cos 2 cos t j cos 6 t i cos3 t sin3 t j 3 cos4 t sin5 t sin4 t 1 cos2 t 2 cos2 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 cos8 t sin t Work C 1 6 cos6 t sin t 2 3 cos4 t sin t 6 cos6 t sin t 3 cos5 t 5 2 0 F dr 0 6 cos8 t sin t 2 cos9 t 3 6 cos7 t 7 3 cos4 t sin t dt 43 105 Section 14.2 32. F x, y yi xj 4 x2 34. F x, y, z yz i xz j xyk Line Integrals 415 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 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 2 sin t j, 2 cos t j 2 cos t j 4 cos2 t 4 0 0≤t≤ 3t j 3j 2 t k, 2k 0≤t≤1 10 t 2 j 15t 2 k 4 cos 2 t Work F C dr 0 90 t 2 d t 30 Work dr cos 2 t d t 2 sin 2 t 0 0 36. 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 38. F x, y (a) x2 y i r1 t r1 t Ft F dr 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 3 2 1 2t2 2 cos t i C1 0 (b) r2 t r2 t Ft F C2 1 4 cos2 t j, 0 ≤ t ≤ 2 sin t i 1 2 8 cos t sin t j 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. 40. F x, y C: r t rt Ft F r C 3y i ti i 3t 3 i 3t 3 F dr t3j 3t 2 j xj 42. F x, y C: r t rt Ft 0 F r 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 tj 3t 3 0. 9 sin t cos t F C Thus, Thus, dr 0. 44. 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 0 202 416 46. x Chapter 14 2t, y 3y C Vector Analysis 5x, dy x dx 2 10t, 0 ≤ t ≤ 1 ⇒ y 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 48. 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 50. 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 52. 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 54. 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 12t 0 5 2 6 Section 14.2 56. f x, y y 58. f x, y C: x2 x y2 rt rt rt 4 Line Integrals 417 y 1 from 1, 0 to 0, 1 cos t i sin t i 1 sin t j, 0 ≤ t ≤ cos t j 2 C: line from 0, 0 to 4, 4 rt rt rt ti i 2 j t j, 0 ≤ t ≤ 4 Lateral surface area: f x, y d s C 0 Lateral surface area: t 2 dt 82 f x, y d s C 0 2 2 cos t sin t d t sin t cos t 0 2 60. f x, y C: y 1 y 1 x2 from 1, 0 to 0, 1 1 i 1 ti 21 41 1 tj t 2 rt rt rt 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 x2 y2 y2 4 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 62. f x, y C: x2 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 0 418 Chapter 14 1 x 4 Vector Analysis 64. 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 66. f x, y C: y S 8 4 x 3 2 179u3 3 z 850,304 91 1215 7184 6670.12 y x 2 from 0, 0 to 2, 4 4 3 2 1 Matches c. 4 y (2, 4, 0) 68. W C F 15 4 15xy dr C M dx 60 15x c 2cx d x 15x 2 c cx2 N dy 15x c cx 2 2 y M N dx W xy 2 cx 2 c y = c ) 1 − x 2) d x, dy 1 60 1 15 x c cx2 2 cx dx −1 x 1 120 w 16c 4c 4 8c 2 (parabola) 0⇒c 1 4 yields the minimum work, 119.5. Along the straight line path, y 72. (a) Work 0 0, the work is 120. 70. See the definition, page 1024. (b) Work is negative, since against force field. (c) Work is positive, since with force field. 74. False, the orientation of C does not affect the form f x, y d s. C 76. False. For example, see Exercise 32. S ection 14.3 Conservative Vector Fields and Independence of Path 419 Section 14.3 2. F x, y (a) r1 t r1 t Ft F C Conservative Vector Fields and Independence of Path y2 i xj (b) r2 w r2 w Fw tj t2 0 x2 ti i t2 dr t j, 0 ≤ t ≤ 4 1 2t ti 4 w2 i 2w i w4 dr 0 w j, 0 ≤ w ≤ 2 j w2 i 2 j w2j w2 w3 3 2 0 t t2 2 t 1 2 3 F t dt 80 3 C 2w w 4 w6 3 w4 2 w 2 dw 80 3 t3 3 4. F x, y (a) r1 t r1 t Ft F C 32 4 0 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 xy k z 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 8. F x, y, z curl F P y x z y ln z i x ln z j 0 so F is not conservative. x z N z M , F is conservative. y sin yz i xz cos yz j x y sin yz k 10. F x, y, z curl F 12. F x, y (a) r1 t r1 t Ft F C 0, so F is not conservative. ye xy i ti i t dr 0 3 xe xy j t j 3e 3 3t t2 3 j, 0 ≤ t ≤ 3 (b) F x, y is conservative since M y te 3t t2 i te 3e 3t t2 j dt N x x ye xy e xy. e xy k. t e 3t 0 3 t2 3t t2 The potential function is f x, y 3 e0 2 t dt e0 0 e 3t t2 0 420 Chapter 14 xy2 i ti i 1 i t dr 1 Vector Analysis 2x2y j 1 j, t 1 j t2 2tj 3 14. F x, y (a) r1 t r1 t Ft F C 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 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 18. C x2 Since M y2 dx y x2 2 xy d y Nx y2 i 2y, 2 xy j 20. F x, y, z i zj yk 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 8, 4 F x, y sin t j x 2t i yz t 2 k, 0 ≤ t ≤ 1, 0, 2 is conservative. The potential function is f x, y (a) C dr 1 2 1, 0, 0 x3 3 y2 dx xy2 2 xy d y k. x3 3 x3 3 xy2 0, 0 0, 2 x2 896 3 8 3 (b) r2 t F C 2 t k, 0 ≤ t ≤ 1 1, 0, 2 dr x yz 1, 0, 0 2 (b) C x2 y 2 dx 2xy dy xy 2 2, 0 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 —CONTINUED— Section 14.3 22. —CONTINUED— (b) r2 t r2 t Ft F C Conservative Vector Fields and Independence of Path 421 1 2i 1 2t i k 2t j 1 t k, 0 ≤ t ≤ 1 3 3 0 2t 2 3t 2 1 1 2t k 1 dr 2t d t 3 3 0 t2 2t 3 dt 3 3 t3 3 t4 2 1 0 3 2 4, 3 24. F x, y, z (a) r1 t r1 t Ft y sin z i t2i 2t i x sin z j x y cos x k 26. C 2x yi 2x yj dr x 49 y 2 3, 2 t 2 j, 0 ≤ t ≤ 2 2tj t 4 cos t 2 k 2 F C dr 0 0 dt 0 (b) r2 t r2 t Ft 4t i 4i 16 t 2 F C 4 t j, 0 ≤ t ≤ 1 4j cos 4 t k 1 dr 0 0 dt 0 28. y dx x2 C 2x C x dy y2 arctan x y 2 3, 2 1, 1 3 1 x2 4 12 1, 5 30. x2 y2 2 dx 2y x2 y2 2 dy y2 7, 5 1 26 1 74 12 481 32. C zy dx xz dy x y dz x yz k, the integral is y z i x z j x y k is conservative and the potential function is f x, y, z Note: Since 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 4, 3, 1 34. C 6x d x 4z d y 4y 20z dz 3x 2 4yz 10z 2 0, 0, 0 46 36. F x, y Work 2x i y x2 y 1, 4 x2 j is conservative. y2 1 4 9 2 17 4 3, 2 422 Chapter 14 a1i Vector Analysis a2 j a3k 38. F x, y, z 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 . 40. 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 42. F x, y (a) M M y N N x Thus, (c) r t F dr y x2 y x2 x2 x x2 x2 N x cos t i sin t i sin t i F C 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 This does not contradict Theorem 14.7 since F is not continuous at 0, 0 in R enclosed by curve C. (e) arctan x y 1 x2 1y i x y2 y y2 i x2 1 x y2 x y2 j x y2 j F 44. A line integral is independent of path if C F d r does not depend on the curve joining P and Q. See Theorem 14.6 46. No, the amount of fuel required depends on the flight path. Fuel consumption is dependent on wind speed and direction. The vector field is not conservative. 48. True 50. False, the requirement is M y N x. Section 14.4 Green’s Theorem 423 Section 14.4 t i, 4i 12 Green’s Theorem y 2. r t t 4 j, ti 12 4 t j, 0≤t≤4 4≤t≤8 8 ≤ t ≤ 12 8 (4, 4) 4 3 2 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 y 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 4. r t cos t i y2 dx C sin t j, 0 ≤ t ≤ 2 2 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. 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 1 ex x3 xe y d y d x 0 xe x x3 e x d x 0.22 In Exercises 8 and 10, N x M y 1. 8. Since C is an ellipse with a y C 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 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 424 Chapter 14 Vector Analysis 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 14. In this case, let y x2 C r sin , x 2xy d y r cos . Then d A 2 r dr d and Green’s Theorem yields 1 0 cos y2 dx 4y dA R 2 1 0 2 cos 4 0 r sin r dr d 4 0 r 2 sin dr d sin 0 4 3 1 cos 42 3 d 1 cos 3 0. 0 16. Since M y N x 2ex sin 2y M dA y N we have x 0. R 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 . 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 22. F x, y C: r Work C e 2 2 2 e2 2e e 2e 1 e 2 16e 2e 1. 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. 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 4 x 0, x 4 4 x1 2 Work C 3x 2 y dx 4 xy 2 d y 0 0 4y2 4 32 3x dx 176 15 S ection 14.4 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 Green ’s Theorem 425 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 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 2t3 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 30. See Theorem 14.9: A 1 x dy 2C y dx. 32. (a) For the moment about the x-axis, Mx R y dA. Let N Mx 2A 1 2A 0 and M y 2 d x. y 2 2. By Green’s Theorem, Mx C y2 dx 2 1 2 y 2 d x and y C C For the moment about the y-axis, My R x dA. Let N My 2A 1 2A x 2 dy. C x 2 2 and M 0. By Green’s Theorem, My x2 dy C2 1 2 x 2 d y and x C (b) By Theorem 14.9 and the fact that x A 1 2 x dy y dx 1 2 r cos r cos , y r cos d r sin , we have r sin r sin d 1 2 r2 d . C 34. Since A Let x 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 x, y 0, 4a 3 426 Chapter 14 1 2a c 2 0, dy c b c b a a Vector Analysis 1 2A 1 , 2ac y 36. Since A C1: y C2: y C3: y Thus, x y ac, we have 0 (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 a2 2 a 3 c . 3 38. A 1 2 a 2 cos 2 3 d 0 1 0 cos 6 d 2 a2 4 sin 6 6 ≤ . 0 a2 4 Note: In this case R is enclosed by r 40. In this case, 0 ≤ u ≤ 2 and we let 1 1 u2 ,d u2 a cos 3 where 0 ≤ sin , cos 1 cos as ⇒ 1 2 du . u2 Now u ⇒ 1 2 2 and we have 9 cos 2du 1 u2 1 u2 1 4 1 u2 1 6 arctan 3 3u 0 A 0 2 2 d 9 0 4 u2 2 u2 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 42. (a) Let C be the line segment joining x1, y1 and x2, y2 . y dy y2 x2 y2 x2 y dx C 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 dx x1 x1 x2 x1 y2 x2 y1 x1 y1 d x x1 x1 y2 —CONTINUED— y1 x y1 x2 x1 y2 x2 y1 x1 x1 y2 x2 y1 Section 14.5 42. —CONTINUED— (b) Let C be the boundary of the region A Therefore, dA R Parametric Surfaces 427 1 2 y dx C x dy 1 2 1 R 1 dA R d A. 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 . 44. Hexagon: 0, 0 , 2, 0 , 3, 2 , 2, 4 , 0, 3 , A 1 2 1, 1 0 0 3 0 0 21 2 0 F C 0 N ds 4 0 12 4 6 46. Since div F d A, then R f DN g ds C C fg N ds f div R div f g d A R g f g dA R f 2g f g d A. 48. C f x dx g y dy R x gy y fx dA R 0 0 dA 0 Section 14.5 2. r u, v x2 y2 u cos v i z2 Parametric Surfaces u sin v j uk 4. r u, v x2 y2 4 cos u i 16 4 sin u j vk Matches d. 12 uk 2 12 x y2 8 Matches a. 6. r u, v z 2u cos v i y2 2 u sin v j 4u2 ⇒ z 8. r u, v x2 x2 9 x2 9 y2 9 y2 y2 3 cos v cos u i 9 cos2 v cos2 u z2 25 z2 25 cos2 v 1 3 cos v sin u j 9 cos2 v sin2 u sin2 v 1 5 sin v k 9 cos2 v 12 2 u,x 2 Paraboloid z 4 Ellipsoid z 4 x 4 y 5 x 4 3 34 y 428 Chapter 14 Vector Analysis z For Exercises 10 and 12, r u, v u cos v i u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 . 5 Eliminating the parameter yields z x2 y 2, 0 ≤ z ≤ 4. 2 x 2 y 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. 12. s u, v z x2 16 y2 64. It was x 2 16. r u, v y2 4. 2 u sin v j v k, z 4 u cos v i y2 4 u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 The paraboloid is “wider.” The top is now the circle x 2 14. r u, v 2 cos v cos u i 4 cos v sin u j z 5 4 3 sin v k, 2u cos v i 0≤u≤2 ,0≤v≤2 x2 4 y2 16 z2 1 1 −5 −4 −5 0 ≤ u ≤ 1, 0 ≤ v ≤ 3 y z arctan x 8 4 5 x 4 3 −3 −4 −5 5 y x 4 −4 −2 2 2 4 y −4 18. r u, v 0≤u≤ cos3 u cos v i 2 ,0≤v≤2 sin3 u sin v j z u k, 20. z 6 x ui y vj 6 u vk r u, v 2 −1 −1 1 1 y x 22. 4x 2 r u, v y2 16 2 cos u i 4 sin u j vk 24. x2 9 r u, v y2 4 z2 1 1 2 cos v sin u j sin v k 3 cos v cos u i 26. z x2 y 2 inside x 2 v cos u i y2 9. v 2 k, 0 ≤ v ≤ 3 28. Function: y x u, y x3 2, 0 ≤ x ≤ 4 u3 2 cos v, z u3 2 sin v r u, v v sin u j Axis of revolution: x-axis 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: y-axis u 2 cos v, y 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 Section 14.5 Parametric Surfaces 12 u k, 2 uk 429 32. r u, v ru u, v ui i vj uv k, 1, 1, 1 j u k 2 uv 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. v k, rv u, v 2 uv 1 and v 1 k, rv 1, 1 2 rv 1, 1 1. j At 1, 1, 1 , u ru 1, 1 i 4, 0, 2 , u 2, 0 ru rv 2i 1 k 2 1 2i 1 2j ru N k 2 k, rv 8i 8k 2, 0 4j N ru 1, 1 ijk 101 2 1 012 2 y 2z 0 1 2z Direction numbers: 1, 0, 1 Tangent plane: x x 4 z z 2 2 0 Direction numbers: 1, 1, Tangent plane: x x 36. r u, v ru u, v rv u, v ru ru A 0 0 1 y 1 0 4 u cos v i 4 cos v i 4 u sin v i 4 u sin v j 4 sin v j u2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 2u k 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 256u2 4 du dv 0 8u 2 cos v i 4 128 2 3 8u 2 sin v j 16 u k 8u u 2 2 64 dv 3 128 3 22 1 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 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 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 k 0 b cos v b cos v j b sin v a b cos v k 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 2ab 430 Chapter 14 Vector Analysis uj j sin u sin v k, 0 ≤ u ≤ cos u sin v k sin u cos v k cos u sin u j cos2 u cos2 u d u d v 22 ln 2 2 1 1 sin u sin v k ,0≤v≤2 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 rv rv 2 1 1 44. See the definition, page 1055. 46. Graph of r u, v 0≤u≤ (a) 10, 0, 0 z 3 −3 u cos v i ,0≤v≤ u sin v j from vk (b) 0, 0, 10 (c) 10, 10, 10 z 3 y y −3 3 3 3 −3 3 x x y 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. 50. x 2 Let x ru u, v rv u, v 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 u sin v i u cos v j. 0. ru 1, 0 is undefined and rv 1, 0 j. The tangent plane at 1, 0, 0 is x 1. At 1, 0, 0 , u 1 and v Section 14.6 52. r u, v ru u, v rv u, v ru ru A 0 a b Surface Integrals 431 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 fuf ui f u cos v j f u sin v k rv rv k i j 1 f u cos v f u sin v 0 f u sin v f u cos v fu 2 b 1 fu 1 fu 2 fu 2 du dv 2 a fx 1 fx 2 dx since u x Section 14.6 2. S: z 15 x S Surface Integrals 3y, 0 ≤ x ≤ 2, 0 ≤ y ≤ 4, 2 4 2x 2y 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 4. S: z 2 32 x , 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 3 1 x z 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 3 2 18 2 18 2 18 2 18 2 2 1 x 32 0 1 5 12 1 x3 0 2 1 x dx 1 1 x 32 0 51 12 3 1 x3 2 1 x 32 0 5 24 1 x1 0 2 1 x dx 52 18 5 24 1 5 24 x 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 2 x x2 x2 x 0 13 ln 42 5 ln 192 3 432 Chapter 14 Vector Analysis 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 8. S: z 1 xy, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 2 4 4 z x 1 y, 2 z y 1 x 2 3904 15 160 5 3 xy dS S 0 0 xy 1 y2 4 x2 dy dx 4 10. S: z cos x, 0 ≤ x ≤ x2 S 2 ,0≤y≤ 2 x2 x 2 2 2xy dS 0 0 x2 2xy 1 sin2 x dy dx 0 x3 4 1 sin2 x dx 0.52 12. S: z x, y, z m a2 kz x2 y2 a z kz dS S 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 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 16. S: r u, v ru rv x S 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 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 2 dr d 0 2 1 1 12 4r2 32 2 sin cos d 65 65 17 17 sin2 12 2 0 0 Section 14.6 20. f x, y, z S: z x2 x2 y2 z2 1 2 Surface Integrals 433 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 x2 y2 2 y x2 y2 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 sin3 3 0 0 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 dz dx 2 3 3 9 x2 x2 9 3 3 9z 12 0 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 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 12 434 Chapter 14 xi y2 36 z2 x2 z 36 yj Vector Analysis zk (first octant) 6 5 4 3 y 26. F x, y, z S: x 2 z 36 y2 36 x x2 x2 + y2 = 62 G x, y, z G x, y, z F G F S x2 y y2 F y2 i 2 36 y2 36 x 2 G dA R 2 0 R 2 y x2 y2 36 6 0 1 y j 2 z 36 x2 k 36 x2 dA x 1 2 3 4 5 6 x2 36 x 2 N dS R 36 y2 y2 36 r dr d 36 r 2 (improper) 108 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 y2 y2 F y2 i a2 a2 G dA R 2 0 2 y x2 y2 −a a x y2 2 j k −a a2 N dS x2 x2 y2 x2 a2 x2 2a 2 dA y2 y2 3x 2 3y 2 a2 x 2 2a 2 y2 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 a2 r2 a2 r 2 a 0 a 32 0 dr d 2 r2 2 2a 2 0 r a2 d r2 dr d 2 a 3 0 2 22 a 3 ad 0 r2 2a 2 0 a2 r2 0 d 3 0 23 ad 3 2a 2 0 30. F x, y, z S: z 1 x x2 z yi y 2, z x2 yj 0 1 k zk G x, y, z G x, y, z F G F S y2 2y j 2x i 2x x N dS y 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 cos 2 sin 1 r dr d 3 4 sin2 4 2 0 3 4 d 3 2 The flux across the bottom z 0 is zero. Section 14.6 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. 36. E S: z E S Surface Integrals 435 34. Orientable yz i 1 xz j x2 N dS xyk y2 E R 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 38. x 2 z y2 ± z2 a2 2 S a2 x2 k dS y2 2k R m 1 a x2 a a2 x x2 dA 2 2 y2 2 a 0 a2 r a2 y x2 r2 2 y2 dr d dA 2k R a2 a2 y 2 0 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 y2 r2 dA a 32 0 2ka 0 r3 a2 r2 dr d (use integration by parts) 2ka r2 a2 23 a2 3 r a2 2 2ka Let u 40. z x2 22 a 4 ka2 3 r2 12 22 am 3 2r dr, v a2 r 2. r 2, dv dr, du 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 x h 2 0 0 x2 2 y 2 1 4x 2 4y dy dx 2 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 1 436 Chapter 14 16 x2 Vector Analysis y2 42. S: z F x, y, z 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 0.5 0 0 16 r 2r dr d 0.5 0 Section 14.7 Divergence Theorem 2. Surface Integral: There are three surfaces to the cylinder. Bottom: z 0 dS S1 0, N 0 k, F k, F N z2 Top: z h, N h 2 dS N z2 4 h2 v k, 0 ≤ u ≤ 2 , 0 ≤ v ≤ h k 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 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 z 2zr dz dr d 0 h x 2 2 y Section 14.7 4. F x, y, z xy i zj x yk 4, z 4 x and the coordinate planes Divergence Theorem 437 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 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 33 xa dx 2 16 a 3 8. Since div F y z y a z, we have a2 x2 a2 x2 0 a 0 a2 x2 y2 2 a 0 0 a 2 a2 r2 div F dV a Q 2 0 z dz dy dx 0 zr dz dr d r4 8 a4 d 8 a4 . 4 a 2r 2 r3 dr d 2 2 0 a 2r 2 4 d 0 0 10. Since div F xz, we have 4 3 3 9 9 y2 4 3 xz dV 0 Q y2 xz dx dy dz 0 z 0 dy dz 2 3 0. 438 Chapter 14 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 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. 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 18. Using the Divergence Theorem, we have curl F S Q N dS i div curlF dV j y yz sin x z k z xyz x sin z 0. a a a a a curl F x, y, z x xy cos z z xz y sin x i yz xy sin z j yz cos x x cos z k. 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 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 24. If F x, y, z Therefore, F S Q Q 22. v 0 0 x dy dz 0 a a 0 a dy dz 0 a a a 2 dz a 3. a3 0, then incompressible. a1 i a2 j a 3 k, then div F 0. Similarly, 0 0 y dz dx 0 0 z dx dy N dS div F d V 0 dV 0. 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 28. S f DN g gD N f dS S f D N g dS S g D N f dS g dV Q f Q 2g f g 2f g f dV Q f 2g g 2f dV Section 14.8 Stokes’s Theorem 439 Section 14.8 2. F x, y, z x2 i i curl F x x 2 Stokes’s Theorem y2 j j y y 2 x2 k k z x 2 4. F x, y, z x sin y i i y cos x j j y y cos x yz2 k k z yz 2 2x j curl F x x sin y z2 i y sin x x cos y k 6. F x, y, z arcsin y i i j y 1 x 1 x 1 1 x2 j k z y2 k curl F x arcsin y 2y i 2y i x2 x2 x2 y2 1 1 1 1 y2 y2 4, z y dx C 2 y2 k k 8. In this case C is the circle x 2 Line Integral: C 0, dz x dy 0. F dr 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 10. Line Integral: From the accompanying figure we see that for C1: y C2: z C3: y C4: z Hence, F C z 1 0, z y, x a, z y 2, x 2 0, dy 0, dx a2, dy a, dx dz 0 2y dy 0 1 C3 0, dz dz 0, dz C1 C2 C4 1 y 2y dy. x dr C z 2 dx 0 C1 a C2 x 2 dy 2y3 dy y 2 dz a4 dx C3 C4 0 0 a 0 a2 dy y2 2y dy a4 x 0 a 2y3 dy 0 0 a4 d x a a2 dy a 2y3 dy a2y a a5 a3 a 3 a2 1. —CONTINUED— 440 Chapter 14 Vector Analysis 10. —CONTINUED— Double Integral: Since F x, y, z N 2yj 1 k and dS 4y 2 2y i 2z j 1 y2 z, we have 4y 2 d A. 2 x k . Therefore, a a a a Furthermore, curl F curl F S N dS R 4yz 2x dA 0 0 4y2 2x dy dx 0 \ a4 2ax d x \ a4 x ax2 0 a 3 a2 1. 12. Let A N 0, 0, 0 , B U U V V x yj 4x y2 1, 1, 1 , and C 2i 2 j 22 y and dS 4 xy k, S: 9 4y j z 2y j 9 k 8x 2 R 3 3 3 9 9 x2 0, 0, 2 . Then U AB i j k , and V AC 2 k, and i j . 2 2 d A. Since curl F x2 y 2, z ≤ 0 2x x2 y2 k , we have S Hence, F x, y, z 14. F x, y, z curl F G x, y, z G x, y, z 4xz i 4x i x2 2x i curl F S curl F N dS R 0 dS 0. N dS 2y 4x 8x 2 x2 4y d A 8xy 16 9 3 x2 y2 8y 2 d y d x x2 32 16x2 9 3 x2 dx 0 16. F x, y, z x2 i i j y z2 z2 j k z x yz k, S: z 4 curl F x x2 xz 2z i yzj xyz 4 x2 x x2 N dS R G x, y, z G x, y, z z 4 curl F y2 y2 i 4 y x2 zx 4 xx R 2 y2 j k y 2z 4 x2 2 4 4 S x2 2 2x y2 y 2 dA y2 x2 dA x2 2x y2 dy dx 2 x2 x2y 2 2 2xy x2 x2 4 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 16 arcsin 4 3 81 32 x4 x2 4 arcsin x 2 2 2 4 2 3 2 Section 14.8 x2 y2 k Stokes ’s Theorem 441 18. F x, y, z yz i i 2 j y 2 3y j k z curl F x yz 2y i y2 z2 y 2x j zk 3y x 2 S: the first octant portion of x 2 G x, y, z G x, y, z z 16 x 16 curl F S 16 over x 2 y2 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 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 0 442 Chapter 14 zi y2 i curl F x z 1 j y 0 Vector Analysis yk 22. F x, y, z S: x 2 k z y N i j Letting N k, curl F 0 and S curl F N dS 0. 24. curl F measures the rotational tendency. See page 1084. 26. f x, y, z (a) x yz, g x, y, z k x yz k 2 sin t j dr xyk 0 k, 0 ≤ t ≤ 2 0 z, S: z 4 x2 y2 g x, y, z f x, y, z g x, y, z rt 2 cos t i 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 Review Exercises for Chapter 14 2. F x, y 5 4 3 2 x −1 −2 −3 −4 −5 2 4 i y 2y j 4. f x, y, z F x, y, z x 2eyz 2xeyz i xeyz 2i x 2ze yz j xz j x 2 ye y z k xyk Review Exercises for Chapter 14 6. Since M y 1 x2 N x, F is conservative. From M Ux y x 2 and N Uy yields U yx h y and U yx g x which suggests that U x, y yx C. 8. Since M y 6y 2 sin 2x N x, F is conservative. From M Ux 2y3 sin 2x and N we obtain U y3 cos 2x h y and U y3 1 cos 2x g x which suggests that h y y3, g x U x, y y3 1 cos 2x C. 10. Since M y M z N z 4x 2z 6y N , x P , x P y 12. Since M y sin z N , x M z y cos z P , x 443 1 x, partial integration U y 3y 2 1 C, and cos 2x , F is not conservative. F is not conservative. 14. Since F (a) div F (b) curl F xy 2 j 2 xy 2 xz j x2 2x 0 (b) curl F z x 2 k; x2 y 2k 16. Since F (a) div F (b) curl F sin2 y j: 20. Since F (a) div F 3x 3 2i z i x z x2 yi 1 3j z j y z y2 1 j x 1 y 5 k z 2 k: 2z 2z j z 3x k: 18. Since F (a) div F (b) curl F yi x 2 sin y cos y z2 1 x2 1 y2 1 i y 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 16 5 t2 2 dx dt 8t 2 2 5 dt 0 0 dt t3 3 1 0 85 . 3 dy dt 2 24. x t sin t, y 2 1 t 0 2 cos t, 0 ≤ t ≤ 2 , sin t 1 cos t 2 1 cos t, 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 8 444 Chapter 14 Vector Analysis 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 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 30. f x, y C: y rt rt rt 12 2 x y x from 0, 0 to 2, 4 ti i t 2 j, 0 ≤ t ≤ 2 2t j 1 4t 2 Lateral surface area: 2 f x, y ds C 0 12 t t2 1 4t 2 dt 41.532 32. dr F F C 4 sin t i 4 cos t 2 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 dr 7 sin t cos t dt 24 0 34. x dr F F C 2 t, y i 4 dr 0 2 j t, z 4t t 2, 0 ≤ t ≤ 2 2t k dt 4t t 2 4t t2 i 2 dt t2 2 4t 2 2t 2 t2 2 2 tj 0k t 2t 0 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 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 Review Exercises for Chapter 14 2000 5280 t k, 0 ≤ t ≤ 2 2 25 tk 33 445 40. r t 10 sin t i 10 sin t i 10 cos t j 10 cos t j F dr F C 20k 10 cos t i 2 10 sin t j 500 dt 33 25 k 33 250 mi 33 ton dr 0 42. C y dx x dy 1 dz z ,y a1 y dx. 4, 4, 4) xy ln z 0, 0, 1 16 ln 4 44. x a 1 2 sin 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 14.4, y 2 dx 1 2A 2 a3 1 0 cos 2 1 cos d C 1 a3 5 2 3 a2 5 a 6 2 2 a a2 a2 x2 46. C xy dx x2 y 2 dy 0 2 0 2x 2x dx 0 x dy dx 4 48. C x2 y 2 dx 2xy dy a a x2 4y dy dx 0 dx a 0 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 0 446 Chapter 14 Vector Analysis u k 6 52. r u, v e u4 cos v i e u4 sin v j 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 z 2 2 x 2 y 54. S: r u, v ru u, v ru 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 sin v 2 cos2 v 0 0 56. (a) z z aa 0 ⇒ x2 z x2 y2 a2 y2 , 0 ≤ z ≤ a2 a2 a x2 y2 a2 z (b) S: g x, y x, y m S k x2 y2 a a y 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 y2 dA 1 0 2 0 r2 dr d a3 d 3 1 0 1 a3 Review Exercises for Chapter 14 447 58. F x, y, z xi yj 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 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 ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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