math192_hw8sp06 - HW8 Solutions 15.3 Double Integrals in...

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Unformatted text preview: HW8 Solutions 15.3: Double Integrals in Polar Form 1 0 - 0 - 10. -1 2 1-y 2 4 x2 + y 2 dxdy = 1 + x2 + y 2 xy 2 dxdy = rdrd = 3/2 /2 0 2 sin 1 4r2 drd = 4 1 + r2 3/2 /2 0 1 1- 1 1 + r2 drd = 4 - 2 14. 0 1-(y-1)2 /2 1+cos 1 1-cos 1 /2 0 0 a /2 0 /2 0 4 sin2 cos r4 drd = - 5 8+ 4 18. A = 2 0 3/2 (2 cos + cos2 )d = 26. I0 = /2 rdrd = 2 + r 4 4 3a2 /2 0 4 29. average = a2 a2 - r2 drd = a3 d = 2a 3 15.4: Triple Integrals in Rectangular Coordinates 2 3 0 0 4-x2 2 3 0 2 4. = 3 x 2 3 0 2 0 0 1 0 3 2 0 dzdydx = 4 - x2 + 4 sin-1 0 0 3-3x 0 1 1-x2 0 2 2 3 0 4-x2 -y 3-3x-y 4-x2 x 2 2 0 2 0 4 - x2 dydx = 0 3 4 - x2 dx = 6 sin-1 1 = 3 0 4-x2 3 2 4-z 2 0 0 3 dzdxdy, 0 4-z 2 3 2 0 0 0 4-z 2 dydzdx, 0 dydxdz, dxdydz, 0 dxdzdy dzdydx = 3 2 1 1-x2 0 0 1 0 10. 0 16. 1 0 0 xdzdydx = x(1 - x2 - y)dydx 1 x (1 - x ) - (1 - x2 ) dx = 2 1 1 x(1 - x2 )2 dx = - (1 - x2 )3 2 12 1 = 0 1 12 22. 1 1 0 1 (a) 0 - z dydzdx dydxdz dxdydz (b) 0 1 -1 1 - z (c) 0 0 -1 - z 1 -1 0 1 0 - 1-x2 4-x2 0 26. V = 2 0 2 -y 0 3-x 1 0 - 1-x2 1 dzdydx = -2 dzdydx = 12 ydydx = 0 0 (1 - x2 )dx = 2 3 32. V = 2 -2 - 4-x2 2 4-x x 0 44. 4 0 = 0 sin 2z 4-z 2 4-x x sin 2z sin 2z dydzdx = dzdx = 4-z 4-z 0 0 0 4 1 1 1 1 (4 - z)dz = - cos 2z = - + sin2 z 2 4 4 2 0 2 4 0 4 0 4-z = 0 sin2 4 2 sin 2z 4-z xdxdz ...
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