ODDREV13 - R eview Exercises for Chapter 13 169 Review...

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Unformatted text preview: R eview Exercises for Chapter 13 169 Review Exercises for Chapter 13 x2 x2 1. 1 x ln y dy xy 1 ln y 1 x3 1 ln x 2 x x x3 x3 ln x 2 1 1 0 x 1 1 x 1 3. 0 3x 2y dy dx 0 3xy y2 0 dx 0 4x 2 5x 1 dx 43 x 3 52 x 2 1 x 0 29 6 3 9 0 x2 3 5. 0 4x dy dx 0 4x 9 x 2 dx 4 9 3 3 x2 32 0 36 3 3 0 x3 1 3 0 3y 7. 0 dy dx 0 1 3 0 3y dx dy 1 A 0 dx dy 0 3 3y dy 3y 32 y 2 3 1 0 3 2 5 25 25 y2 3 25 25 x2 4 25 25 y2 4 9. 5 x2 3 dy dx 5 25 x2 y2 3 dx dy 4 25 y2 dx dy 4 y2 dx dy x 5 3 5 A 2 50 dy dx 2 5 25 x2 dx x 25 x2 25 arcsin 25 2 12 25 arcsin 3 5 67.36 1 x 0 1 x2 1 11. A 4 0 12 1 1 dy dx 1 1 4y 2 4y 2 4 0 2 x1 x2 dx 4 1 3 1 x2 32 0 4 3 A 4 0 2 dx dy 5 x x 3 1 2 x 0 1 2 y y2 3 13. A 2 dy dx 2 1 dy dx 1 1 dx dy 9 2 y 15. Both integrations are over the common region R shown in the figure. Analytically, 1 0 2 0 2 2y x2 2 2 y2 x y dx dy 4 3 2 0 4 3 8 2 y= 1x 2 2 1 (2, 1) y=1 2 8 − x2 x x2 2 x 0 y dy dx 2 x y dy dx 5 3 4 3 2 1 3 4 3 4 3 2 −1 1 2 3 4 x2 4 17. V 0 4 0 x2 12 y 2 4x2 43 x 3 y 4 dy dx x2 4 19. Volume base height 9 3 2 27 2 6 4 2 z x2y 0 4 0 4y 0 dx Matches (c) 8 dx 4 14 x 2 y 4 2 (3, 3) 4 x (3, 0) 15 x 10 8x 0 3296 15 170 Chapter 13 Multiple Integration 21. 0 0 k xye Therefore, k 1 x y dy dx 0 k xe x y y 1 0 dx 0 k xe x dx kx 1e x 0 k 1. 1 P 0 0 x ye x y dy dx 0.070 23. True 25. True h x 4 h sec 27. 0 0 x2 y 2 dy dx 0 0 4 r2 dr d h3 3 h3 sec tan 6 4 sec3 d 0 ln sec tan 0 h3 6 2 ln 2 1 h 2 0 2 1 1 z2 29. V 4 0 h r dr d dz 31. (a) x2 r2 2 y2 2 9 x2 y2 r 2 sin2 sin2 9 cos 2 9 r 2 cos2 9 cos2 2 0 h 0 1 z2 1 d dz r2 r 3 cos 2 4 z 2 dz 0 13 z 3 h 0 h3 3 −6 6 −4 4 3 0 cos 2 (b) A 4 0 4 3 0 cos 2 r dr d 9 r 2 r dr d (c) V 4 0 9 20.392 1 2x 33. (a) m Mx My x y y k 0 1 2x 3 xy dy dx 2x k 4 16k 55 8k 45 1 2x (b) m Mx My x y k 0 1 2x 3 x2 2x y2 dy dx y2 dy dx y2 dy dx 17k 30 392k 585 156k 385 k 0 1 2x 3 xy2 dy dx 2x k 0 1 2x 3 y x2 2x k 0 2x 3 x2y dy dx 32 45 64 55 k 0 2x 3 x x2 936 1309 784 663 My m Mx m y = 2x My m Mx m 2 1 y = 2x 3 x 1 2 Review Exercises for Chapter 13 171 a b 35. Ix R y2 x, y dA 0 a 0 b kxy2 dy dx 1 32 kb a 6 1 kba4 4 3a2 37. S R 4 1 16 0 2 4 x2 fx 2 fy 2 dA Iy R x2 x, y dA 0 0 kx3 dy dx 1 kba 4 4 a b 4 0 1 4x2 4y2 dy dx I0 m Ix Iy 1 32 kb a 6 ka2b 2 2b 12 1 kba2 2 a2 2 b2 3 4 0 0 1 13 65 3 4r 2 r dr d 2 x, y dA R 0 0 kx dy dx 1 4 kba 4 1 2 kba2 1 6 kb3a2 1 2 kba2 2 1 0 6 65 65 1 x Iy m Ix m 9 y2 2y 1 R 3 0 3 y a2 2 b3 3 y 39. f x, y fx S 0, fy fx2 1 fy2 dA 4y2 dx dy y y 1 0 3 4y2 x y dy 12 1 43 3 21 0 4y2 dy 4y2 32 0 1 37 6 32 1 3 9 9 x2 9 2 3 0 3 r 9 41. 3 x2 x2 y2 x2 y 2 dz dy dx 0 2 0 2 r2 dz dr d 2 9r 2 0 r 4 dr d 0 3r 3 r5 5 3 d 0 162 5 2 d 0 324 5 a b 0 c a b 0 43. 0 0 x2 y2 z2 dx dy dz 0 a 0 13 c 3 cy 2 13 bc 3 cz2 dy dz 1 abc 3 3 13 ab c 3 13 a bc 3 1 abc a 2 3 13 bc 3 bcz 2 dz b2 c2 1 1 1 x2 x2 1 1 x2 y2 2 1 0 1 1 r2 r2 45. 1 x2 x2 y2 y 2 dz dy dx 0 r 3 dz dr d 8 15 172 Chapter 13 2 2 cos 0 2 2 cos 0 Multiple Integration 4 r2 47. V 4 0 r dz dr d 4 0 2 0 0 r4 4 4 3 1 0 r 2 dr d 2 cos r2 32 0 d 32 3 32 3 2 sin3 d 1 cos3 3 2 0 cos 32 32 2 3 2 2 0 cos 2 49. m 4k 4 0 2 sin ddd dd sin 2 k 3 ddd 1 k 2 2 2 4 k 3 Mxy 4k 2 4 2 4 2 0 2 0 cos3 2 cos sin 3 cos3 4 sin d 2 k 3 1 cos4 4 2 4 k 24 cos 0 k 4 0 cos5 k k 96 24 sin 1 4 dd cos5 4 sin d 1 k cos6 12 2 4 k 96 z x Mxy m y 0 by symmetry 2 2 0 2 2 0 a 2 0 0 a 51. m Mxy x k k 0 0 sin cos ddd 2 k a3 6 ddd 3a 8 z 2 4 3 16 0 r2 53. Iz k a4 16 4k 0 2 4 r3 dz dr d 16r 3 0 3 sin 4k r 5 dr d 833 k 3 y z Mxy m a2 a2 k a4 16 6 k a3 55. z f x, y x2 r2 h2 y2 a−h a 0≤r≤ h 2ah x a a y (a) Disc Method a V a h a2 a2y a3 y3 3 a3 3 y2 dy a y a3 a h a3 3 a3 3 a2h a2 a ah2 h h3 3 a 3 h 3 2a a y = a2 − x2 a3 a2h ah2 h3 3 12 h 3a 3 h a−h −a x a Equivalently, use spherical coordinates 2 cos 0 1 a ha a 2 V 0 a h sec sin ddd —CONTINUED— Review Exercises for Chapter 13 55. —CONTINUED— 2 cos 0 1 173 a ha a (b) Mxy 0 a h sec cos 12 h 4 Mxy V 2a h 2 2 sin ddd z 12 h 2a 4 12 h 3a 3 0, 0, 3 2a 4 3a h h h2 h 3 a 8 2 3 2a 4 3a h2 h centroid: (c) If h a, z 3a2 4 2a centroid of hemisphere: (d) lim z h →0 3 0, 0, a 8 3 4a2 12a a h →0 2 cos 0 lim 3 2a 4 3a sin2 1 h2 h (e) x2 Iz y2 2 0 a ha a 2 a h sec sin2 2 sin ddd h3 20a2 30 (f ) If h a, Iz 15ah a3 20a2 30 3h2 15a2 3a2 45 a 15 2 6 sin 2 57. 0 0 0 sin ddd 59. x, y u, v 1 xy uv 3 yx uv 23 9 6 sin represents (in the yz-plane) a circle of Since radius 3 centered at 0, 3, 0 , the integral represents the volume of the torus formed by revolving 0 < <2 this circle about the z-axis. x, y u, v x 1 u 2 xy uv v ,y xy vu 1 u 2 1 2 1 2 11 22 x y, v 1 2 x y y 61. 3 y=x+1 y = −x + 5 v ⇒u 2 Boundaries in xy-plane x x x x y y y y ln x R Boundaries in uv-plane u u 3 5 1 1 v 3 ln u 1 u 2 3 v 1 2 5 ln 5 5 1 1 y = −x + 3 y=x−1 x 3 5 1 1 5 1 1 2 3 v v y dA 3 1 ln 5 ln 5 1 u 2 5 dv du 3 1 ln u dv du 12 2.751 5 5 ln u du 3 u ln u u 3 3 ln 3 2 174 Chapter 13 Multiple Integration Problem Solving for Chapter 13 z 1. (a) V 16 R 4 1 1 x2 dA 1 16 0 0 4 0 1 1 cos2 r 2 cos2 r dr d 1 cos cos2 tan 0 32 16 3 1d 1 1 y R y=x 16 sec 3 82 2 4 x 4.6863 (b) Programs will vary. 3. (a) du a2 Then u2 2 22 1 u arctan a a 1 u2 2 2 u2 v2 dv c. Let a2 1 2 v 2 u 2 u 2 u2 u2 u2 u2 2 u2, u v 2 v. C. arctan u u2 (b) I1 0 22 0 22 0 arctan du u 2 2 4 2 u2 u2 arctan arctan u 2 u2 du arctan du u2 2 2 sin2 2 cos2 . Let u I1 4 0 2 sin , du 6 2 cos d , 2 2 sin 2 cos 42 2 v 2 u 2 2 2 u2 u 6 1 2 cos arctan tan arctan 2 cos d 2 2 6 4 0 2 d 2 0 u 6 2 18 du (c) I2 22 2 22 2 22 2 2 2 2 4 2 arctan u2 arctan 2 u2 2 u2 u du u2 arctan u 2 2 u2 du u2 arctan Let u I2 4 6 2 sin . 2 1 2 cos arctan 1 arctan sin cos 2 2 sin 2 cos 2 cos d 2 4 6 d —CONTINUED— Problem Solving for Chapter 13 3. —CONTINUED— (d) tan 1 22 1 1 cos cos 2 2 2 175 1 1 1 sin sin sin cos2 2 1 2 1 sin sin 1 d 2 1 sin 2 sin cos (e) I2 4 6 arctan 2 1 sin cos d 4 6 arctan tan 1 22 d 4 6 1 22 2 2 2 6 2 2 2 d 2 2 2 2 18 2 96 72 1 xy 2 6 4 2 8 4 36 ... 2 12 2 72 2 1 1 1 0 1 9 xy < 1 . . . dx dy 1 (f ) xy 1 0 xy 1 2 1 1 xy 1 dx dy 0 1 0 0 1 1 xy xy 2 xy K dx dy 0K 1 K 0 0 0 K 0 0 xK K 1 2 0 1 K1 y 1 dy 0 yK K 1 1 dy K 0 yK 1 K1 K 0 K 1 2 n 1 n2 1 y (g) u u u x y ,v 2 y x 2 u 2 u 2 1 1 2 2 1 v 2 1 v v x, y u, v 2x ⇒x 2 2y ⇒y 2 1 1 S 0, 0 1 , 2 2 2 v R v x R 0, 0 ↔ 1, 0 ↔ 0, 1 ↔ 1, 1 ↔ 1 0 1 0 ( 1 , 2 1 2 ) ( 2, 0) u 3 4 S 1 2 1 −1 −2 2 1 1 , 2 2 2, 0 22 u u ( 1 −1 , 2 2 ) 1 1 xy dx dy 0 1 2 1 u2 2 2 2 u 2 2 v2 2 9 dv du 22 u 1 1 u2 2 v2 2 dv du 2 I1 I2 18 6 176 Chapter 13 Multiple Integration Boundary in uv-plane u u v v 1 2 3 3 x 2 5. Boundary in xy-plane y y y y x 2x 12 x 3 12 x 4 x, y u, v 1 3 2 3 v u v u 23 7. 6 z (3, 3, 6) 5 4 (0, 0, 0) 4 1 3 (3, 3, 0) (0, 6, 0) 6 y 13 2 3 1 3 1 u v u v 13 3 2x 0 6 x x V 0 dy dz dx 18 23 A R 1 dA S x, y dA u, v 1 3 9. From Exercise 55, Section 13.3, e Thus, 0 x2 x2 2 dx dx 2 2 and 2 1 xe 2 x2 0 e x2 2 e 0 x2 dx x2 2 1 22 x2e 0 dx 1 2 e 0 dx 4 11. f x, y ke 0 x ya x ≥ 0, y ≥ 0 elsewhere ke 0 0 xa x ya 13. A l w ∆x cos θ x cos P y sec xy f x, y dA dx dy ∆y k 0 e dx 0 e ya dy θ ∆x ∆y These two integrals are equal to b e 0 xa dx b→ lim ae xa 0 Area in xy-plane: a. xy Hence, assuming a, k > 0, you obtain 1 ka2 or a 1 k . ...
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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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