math192_hw12sp06 - HW12 Solutions 16.5 Surface Area and...

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Unformatted text preview: HW12 Solutions 16.5 Surface Area and Surface Integrals 2. p = k, f = 2xi + 2yj - k |f | = 2 0 1 2 12 (4r |f | dA R |f p| S = + R 3/2 6 1) d 2 = 4x2 + 4y 2 + 1 and |f p| = 1; x2 + y 2 6 2 2 6 2 2 + 4y 2 + 1dxdy = 2 + 1rdrd = 4x 4r 4r + 1rdrd = R 0 2 2 49 6 d 0 = = 49 3 2 2 dA = 2 6. S = R 2z R 8 g(x, y, z)d = + 2 2 14. 3 wedge |f | dA = |f p| 1 R 2- (x2 + y2) dA = 2 1 0 2 1 rdrd = 2 2 - 2 2 2-r 17. f (x, y, z) = 2x + 2y + z = 2 f = 2i + 2j + k and g(z, y, z) = x + y + (2 - 2x - 2y) = 2 - x - y p = k, |f | = 3 and |f p| = 1 d = 3dydx; z = 0 2x + 2y = 2 y = 1 - x 3 S (2 - S gd = 1 3 x2 2 - 2x + 2 0 x - y)d = 3 dx = 2 a z 1 1-x (2 0 0 - x - y)dydx = 3 1 0 (2 - x)(1 - x) = 1 2 (1 - x) 2 dx = 24. Flux = R (za) dxdy = R 1 a2 dxdy = a2 (Area of R) = 4 a2 28. g(x, y, z) = x2 +y 2 -z = 0 g = 2xi+2yj-k |g| = 4x2 + 4y 2 + 1 = 4 (x2 + y 2 ) + 1 2 2xi+2yj-k 8x2 n= 2 2 F n = +8y 2-2 ; p = k |g p| = 1 d 4 (x2 + y 2 ) + 1dA Flux 2 4(x +y )+1 8x 2 4(x +y )+1 = R +8y -2 4(x2 +y 2 )+1 2 4 (x2 + y 2 ) + 1dA = 2 0 1 0 R 8x2 + 8y 2 - 1 dA; z = 1 and x2 + y 2 = z x2 + y 2 = 1 Flux = 8r2 - 2 rdrd = 2 3 3 int3 z dxdy = 0 -3 3 3 S yd = -3 0 y 34. f = 2yj + 2zk |f | = 4y 2 + 4z 2 = 6 3 z 0 d = 2z dA = z dA; M = S 1d = X zd = 3 3 -3 0 z 3 z ddxdy = 54; Mxz = 3x 9-y 2 4(y 2 + z 2 ) = 6; p = k |f k| = 2z since 3 3 3 dxdy = 9; Mxy = -3 0 9-y 2 3 z dxdy = 0; Myz = S xd = 3 3 -3 0 dxdy = 27 2 . Therefore, x = ( 27 2 ) 3 3 -3 0 3y 9-y 2 dxdy = = 6 9 3 = 2 , y = 0, and z = 54 9 16.6 Parametrized Surfaces 4. In cylindrical coordinates, let x = r cos , y = r sin , z = 2 x2 + y 2 z = 2r. Then r(r, ) = (r cos )i + (r sin )j + 2rk for 0 r 6, 0 2 14. (a) In a fashion similar to cylindrical coordinates, but working in the xz-plane instead of the xy-plane, let x = u cos v, z = u sin v where u = x2 + z 2 and v is the angle formed by (x, y, z), (y, 0, 0), and (x, y, 0) with vertex (y, 0, 0). Since x - y + = 2 y = x + 2z - 2, then 2z r(u, v) = (u cos v)i + (u cos v + 2u sin v - 2)j + (u sin v)k, 0 u 3 and 0 v 2 (b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xyplane, let y = u cos v, z = u sin v where u = y 2 + z 2 and v is the angle formed by (x, y, z), (x, 0, 0), and (x, y, 0) with vertex (x, 0, 0). Since x - y + 2z = 2 x = y - 2z + 2, then r(u, v) = (u cos v - 2u sin v + 2)i + (u cos v)j + (u cos v)k, 0 u 2 and 0 v 2 2 2 x +y r = 3 , 3 r 4 and 0 2. then, 20. Let x = r cos and y = r sin z = 3 r 1 r(r, ) = (r cos )i + (r sin )j + 3 k rr = (cos )i + (sin )j + 3 k and r = (-r sin )i + i j k 1 1 1 cos sin = - 3 cos i - 3 r sin j + r cos2 + r2 sin2 k = (r cos )j rr r = 3 -r sin r cos 0 1 - 3 cos i - 2 0 1 3 r sin j + 4 r 10 drd = 7 3 10 3 3 rk |rr r | = 1 2 9r 1 cos2 + 9 r2 sin2 + r2 = 10r 2 9 = r 10 3 A= 24. Let the parametrization be r(r, ) = (r cos )i + (r sin )j + r2 k, 1 r 2, 0 2 A = 2 2 2 r 4r + 1drd = 17 17 - 5 5 6 0 1 30. Let the parametrization be r(, ) = (a sin cos )i + (a sin sin )i + (a cos )k, 0 2 S G(x, y, z)d = 2 0 /2 0 2, a cos 2 2 a sin dd = 2 2 4 3 a 0 34. Let the parametrization be r(, ) = (2 sin cos )i + (2 sin sin )j + (2 cos )k, 0 /4, 0 2 r = (2 cos cos )i + (2 cos sin )j - (2 sin )k and r = (-2 sin sin )i + i j k (2 sin cos )j r r = 2 cos cos 2 cos sin -2 sin = (4 sin2 cos )i+(4 sin2 sin )j+ -2 sin sin 2 sin cos 0 4 2 + 16 sin4 sin2 + 16 sin2 + cos2 = 4 sin ; y = (4 sin cos )k |r r = 16 sin cos 2 sin sin and z = 2 cos H(x, y, z) = 4 cos sin sin 2 /4 2 /4 H(x, y, z)d = 0 0 (4 cos sin sin )(4 sin )dd = 0 0 16 sin2 cos sin dd = 0 S 42. Let the parametrization z 2) and 0 2 i j rr r = -r sin r cos cos sin be r(r, ) = (r cos )i + (r sin )j + 2rk, - r 1 (since 0 rr = (cos )i + (sin )j + 2k and r = (-r sin )i + (r cos )j k r r 0 = (2r cos )i + (2r sin )j - rk F nd = F |rr r | |rr r 2 r |ddr = (2r3 sin2 cos + 4r3 cos sin + r)ddr since F = (r2 sin2 )i + (2r2 cos )j - k S F 2 1 2 1 1 nd = 0 0 (2r3 sin2 cos + 4r3 cos sin + r)drd = 0 2 sin2 cos + cos sin + 2 d = 1 6 sin3 + 1 2 sin2 + 1 2 2 0 = ...
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