# EVNREV13 - 400 Chapter 13 Multiple Integration 28. x 4u x,...

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Unformatted text preview: 400 Chapter 13 Multiple Integration 28. x 4u x, y, z u, v, w v, y 4 0 1 4v 1 4 0 w, z 0 1 1 u 17 w 26. See Theorem 13.5. 30. x r cos , y x, y, z r, , z cos sin 0 r sin , z r sin r cos 0 z 0 0 1 1 r cos2 r sin2 r Review Exercises for Chapter 13 2y 2. y x2 y 2 dx x3 3 2y xy 2 y 10y3 3 2 2x 2 2x 2 4. 0 x2 x2 2y dy dx 0 x 2y y2 x2 dx 0 4x 2 2x3 2 x 4 dx 43 x 3 14 x 2 25 x 5 2 0 88 15 3 2 2 4 4 y2 3 6. 0 y2 dx dy 2 0 4 y 2 dy y4 y2 4 arcsin y 2 3 3 0 4 3 2 x 3 6 0 2x 2 6 y y2 8. 0 0 dy dx 2 2 6 y 2 y2 dy dx 0 dx dy A 0 dx dy 1 2 1 6y 2 1 1 y 6 0 3y dy 32 y 2 8 2 3 0 4 6x x2 x2 0 1 1 3 1 9 y 9 3 3 9 9 y 10. 0 2x dy dx 11 4 6x x2 x2 y 4 dy dx 0 y dx dy 8 y dx dy 4 0 A 0 2x dy dx 0 8x 2x2 dx 4x2 23 x 3 64 3 2 y2 1 1 2 5 2 12. A 0 0 dx dy 0 0 dy dx 1 x 1 dy dx 14 3 1 1 x 3 2y y y2 0 1 x 1 x 1 1 1 14. A 0 dx dy 3 dy dx 0 x dy dx 9 2 16. Both integrations are over the common region R shown in the figure. Analytically, 2 0 3 0 5 y y ex 3y 2 2x 3 y dx dy 2 5 5 85 e 5 5 0 x 5 4 3 (3, 2) ex 0 y dy dx 3 ex y dy dx 35 e 5 e3 2 5 e5 e3 85 e 5 2 5 2 1 x 1 2 3 4 5 Review Exercises for Chapter 13 401 3 x 18. V 0 3 0 x y dy dx 12 y 2 x 20. Matches (c) 3 z xy 0 dx 0 2 1 y 2 3 2 3 x2 0 dx 2 x 13 x 2 3 0 27 2 1 x 1 22. 0 0 k xy dy dx 0 1 0 k xy 2 2 kx 3 dx 2 1 0 x 1 1 2 2 dx 0 24. False, 0 0 x dy dx 1 1 x dy dx kx 4 8 Since k 8 0.5 k 8 8. 0.03125 1, we have k 0.25 P 0 0 8xy dy dx 1 1 0 26. True, 0 1 1 x2 1 1 0 dx dy < y2 1 1 x2 dx dy 0 4 4 16 0 y2 2 4 2 28. 0 x2 y 2 dx dy 0 0 r 3 dr d 0 r4 4 4 2 d 0 0 64 d 32 2 R 30. V 8 0 b 2 R2 8 3 r 2r dr d R 32. tan 12 13 8 13 3 ⇒ 2 0.9828 R2 0 r2 2 32 b d The polar region is given by 0 ≤ r ≤ 4 and 0 ≤ ≤ 0.9828. Hence, arctan 3 2 4 82 R 3 4 3 R2 b2 32 0 d y r cos 0 0 r sin r dr d 288 13 b2 32 (8/ 13, 12/ 13) 4 3 2 1 x 2 + y 2 =16 y=2 x 3 θ x 1 8/ 13 4 402 Chapter 13 Multiple Integration 2 2 L h2 2 0 h2 2 0 L xL xL 34. m Mx k 0 L xL xL 2 2 dy dx y dy dx 0 2 kh 2 L 2 0 x L x dx L2 2 7khL 12 h y 2 x y = h 2 − L − x2 2 L k kh 8 2 0 L ( ( x x L 4x L 2x 2 L xL x2 2 dx L2 3x2 L2 x3 L2 xL 2 2 L kh2 8 kh2 8 L 4 0 2x3 L3 x4 2L3 x dy dx x4 dx L4 x5 5L4 L 0 4x kh2 8 17L 10 17kh2L 80 h2 2 0 L My k 0 kh 2 x y My m Mx m 2x 0 x2 L x3 dx L2 5L 14 kh 2 x 2 x3 3L x4 4L2 L 0 kh 2 5L2 12 5khL2 24 5khL2 24 17kh2L 80 12 7khL 12 7khL 51h 140 2 2 4 0 4 0 x 36. Ix R y2 x2 R x, y dA 0 2 x 2 ky3 dy dx kx2y dy dx 0 16,384 k 315 512 k 105 512 k 9 Iy I0 m R x, y dA 16,384k 315 2 Ix Iy 512k 105 4 0 x 2 17,920 k 315 x, y dA 0 ky dy dx 512k 105 128k 15 16,384k 315 128k 15 y2 4 7 128 k 15 x y Iy m Ix m 128 21 38. f x, y R fx 1 S 16 x x, y : 0 ≤ x ≤ 2, 0 ≤ y ≤ x 1, fy fx 2 2 0 2 2y fy 2 y 2 2 4y2 2 4y2 dx dy 0 22 4y2 y2 1 2 12 4y2 dy 2 1 2y 2 2 1 4 18 2 62 ln 4 4y2 2 ln 4 2 ln 2y 18 92 2 2 4y2 4y2 ln 2 32 0 1 18 18 12 ln 2 2 6 22 12 ln 2 2 3 32 52 3 R eview Exercises for Chapter 13 403 40. (a) Graph of f x, y z 25 1 over region R z 50 (b) Surface area R 1 fx x, y 2 fy x, y 2 dA e x2 y 2 1000 cos2 x2 y 2 1000 Using a symbolic computer program, you obtain surface area 4,540 sq. ft. R 50 x 50 y 2 4 4 x2 x2 (x2 0 y2) 2 2 2 0 r2 2 42. 2 x2 y 2 dz dy dx 0 0 r3 dz dr d 1 2 2 0 2 r 5 dr d 0 16 3 2 d 0 32 3 5 25 0 x2 0 25 x2 y2 44. 0 1 1 x2 y2 z2 2 2 0 5 0 2 dz dy dx 0 2 0 2 0 2 1 2 sin ddd 5 arctan 0 sin dd 2 5 0 arctan 5 cos 0 d 2 5 arctan 5 2 4 0 x2 0 4 x2 y2 46. 0 xyz dz dy dx 4 3 2 2 sin 0 16 0 r2 2 2 sin 48. V 2 0 2 r dz dr d 2 0 2 0 r 16 r 2 dr d 2 0 32 sin2 2 sin 2 4 sin4 13 sin 4 d cos 8 0 8 sin2 31 42 sin4 d 2 0 84 1 sin 2 4 29 2 2 a 0 cr sin 2 a 50. m Mxz Mxy x y z 2k 0 2 0 cr sin a 0 2 a 0 r dz dr d 2kc 0 0 r 2 sin dr d 2 a 2 kca3 3 dr d 2 sin d 0 2 kca3 3 sin2 d 1 kca4 8 1 k c2a4 16 2k 0 0 cr sin r 2 sin dz dr d rz dz dr d 0 0 2kc 0 2 a 0 r3 sin2 r 3 sin2 dr d 1 kca4 2 2 2 0 2k 0 Mxz m Mxy m kca4 8 2kca3 3 kc2 0 0 1 24 kc a 4 sin2 0 d 3a 16 3 ca 32 kc2a4 16 2kca3 3 404 Chapter 13 Multiple Integration 52. m 500 3 500 3 2 0 2 3 0 3 0 3 0 2 0 4 25 r2 r dz d dr 1 25 3 r2 32 500 3 3 3 0 2 r 25 0 r2 18 4r d dr 125 3 500 3 14 3 162 2r 2 0 500 3 2 64 3 x Mxy y by symmetry 4 2 5 3 25 25 r2 2 3 zr dz dr d 0 2 0 25 r2 0 r2 zr dz dr d 0 3 0 2 0 2 0 8 81 8 1 25 2 81 4 r2 r dr d 0 13 r 2 9 r dr d 2 1 8 14 r 8 92 r 4 d 0 z Mxy m 81 1 4 162 2 a 2 54. Iz k 0 0 0 sin2 2 sin ddd 4k a6 9 56. x 2 y2 z2 a2 x2 Q 2 1 0 r2 1 y2 d V z2 z2 a2 a2 1 1 y 2 z2 y 2 z2 a2 58. 0 0 r dz dr d Since z 1 r 2 represents a paraboloid with vertex 0, 0, 1 , this integral represents the volume of the solid below the paraboloid and above the semi-circle 4 x2 in the xy-plane. y Iz a a 1 1 x2 a2 y 2 dx dy dz 8 a 15 x, y u, v xy uv 2u 2v yx uv 2u 2v 8uv y 60. 62. x x, y u, v u, y xy uv v ⇒u u xy vu x, v 1 1 u xy 0 1 u 5 4 x=1 3 2 y= 1 x x=5 Boundary in xy-plane x x xy xy 1 1 5 1 5 x dA x2y2 5 1 5 1 Boundary in uv-plane u u v v 1 1 5 1 5 2 1 x 1 y= 1 x 4 5 R u u2 v u 5 1 1 du dv u 5 1 5 1 1 1 v2 5 du dv 1 4 1 v2 dv 4 arctan v 4 arctan 5 Problem Solving for Chapter 13 405 Problem Solving for Chapter 13 1 d c 2 5 4 2 2. z fx 1 S ax by b c Plane 4. A: 0 r 16 10 r2 r dr d 160 r2 r dr d 160 1333 960 523 960 4.36 ft3 1.71 ft3 a ,f cy fx2 fy2 1 R B 0 9 r 16 1 a2 c2 c 2 a2 c2 b2 c2 The distribution is not uniform. Less water in region of greater area. In one hour, the entire lawn receives 2 0 10 0 b2 dA c2 dA R a 2 b c 2 r 16 r2 r dr d 160 125 12 32.72 ft3. a2 b2 c c2 AR 2 2 0 2 4 0 8 r2 6. (a) V 0 2 2 2 r dz dr d 2 0 2 sec 8 42 3 ddd 5 5 (b) V sin 8 42 3 8. Volume 5 6 5 54 84 m3 10. Let v ev 1 ln 1 ,x x 1 , dv x e v, dx 0 dx . x e v v, 2u du ue 0 v dv v ln 1 x dx 0 e dv 0 ve v dv Let u 1 v, u2 dv. u2 ln 1 x dx 0 2u du 2 0 u2e u2 du 2 4 2 (PS #9) 12. Essay 14. The greater the angle between the given plane and the xyplane, the greater the surface area. Hence: z2 < z1 < z4 < z3 ...
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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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