20esolns5

20esolns5 - 1 2 = 1 2 . 6.2.19 As we are integrating over a...

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Homework 5 Solutions MTH 20E 4.4.2 div F = ∂x ( yz ) + ∂y ( zx ) + ∂z ( xy ) = 0 . 4.4.7 div F = ∂x ( y ) = 0 . 4.4.8 div F = ∂x ( - 3 x ) + ∂y ( - y ) = - 4 . 4.4.26 If F were a gradient field, then ∇× F = 0. However, ∇× F = ± ∂y (0) - ∂z ( - 2 xy ) ² i + ± ∂z ( x 2 + y 2 ) - ∂x (0) ² j + ± ∂x ( - 2 xy ) - ∂y ( x 2 + y 2 ) ² k = - 4 y k ± = 0 . 5.2.3 V = ³ 1 0 ³ 1 0 xy dx dy = ³ 1 0 y 2 dy = 1 4 . 5.2.4 Another way to describe the region bounded is 0 x 1 , 0 y 1 , 0 z x 2 + y 4 Hence, our volume is given by V = ³ 1 0 ³ 1 0 x 2 + y 4 dy dx = ³ 1 0 x 2 + 1 5 = 1 3 + 1 5 = 8 15 . 6.2.1 If D is given by x 2 + y 2 1, then D * is given by r 1 , 0 θ 2 π. Then ³ ³ D e ( x 2 + y 2 ) dx dy = ³ 2 /pi 0 ³ 1 0 re ( r 2 ) dr dθ = 1 2 ³ 2 /pi 0 e - 1 = π ( e - 1) . 1
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6.2.2 The region D * can be obtained by inverting the change of variables and applying the inversion to D ; that region is D * : v u 1 - v, 0 v 1 2 . Also, ( x, y ) ( u, v ) = ± ± ± ± 1 1 1 - 1 ± ± ± ± = - 2 Hence, ² ² D ( x + y ) dx dy = ² 1 2 0 ² v 1 - v 2 u du dv = ² 1 2 0 (1 - v ) 2 - v 2 dv = ² 1 2 0 1 - 2 v dv = ³ v - v 2 ´
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Unformatted text preview: 1 2 = 1 2 . 6.2.19 As we are integrating over a cylinder, this problem lends itself nicely to cylindrical coordinates; hence D x 2 + y 2 + z 2 dx dy dz = 3-2 2 2 ( r 2 + z 2 ) r dr d dz = 3-2 2 4 + 2 z 2 d dz = 2 3-2 4 + 2 z 2 dz = 2 4 z + 2 z 3 3 3-2 = 100 3 . 6.2.20 Since the function is even, e-4 x 2 dx = 1 2 - e-4 x 2 dx = 1 2 lim b b-b e-4 x 2 dx = 1 4 lim b 2 b-2 b e-u 2 du ( u = 2 x ) 2 = 4 . 3...
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This note was uploaded on 04/07/2008 for the course MATH 20E taught by Professor Enright during the Fall '07 term at UCSD.

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20esolns5 - 1 2 = 1 2 . 6.2.19 As we are integrating over a...

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