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prac3Asol - = 2 y so Z Z R 2 y dA = Z 1 Z x 3 2 y dydx = Z...

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18.02 Practice Exam 3 A – Solutions 1. a) The area of the triangle is 2, so ¯ y = 1 2 Z 1 0 Z 2 - 2 y 2 y - 2 y dxdy. b) By symmetry ¯ x = 0. 2. δ = | x | = r | cos θ | . I 0 = ZZ D r 2 δ rdrdθ = Z 2 π 0 Z 1 0 r 2 | r cos θ | rdrdθ = 4 Z π/ 2 0 Z 1 0 r 4 cos θdrdθ = 4 Z π/ 2 0 1 5 cos θdθ = 4 5 3. a) N x = 6 x 2 + by 2 , M y = ax 2 + 3 y 2 . N x = M y provided a = 6 and b = 3. b) f x = 6 x 2 y + y 3 + 1 = f = 2 x 3 y + xy 3 + x + c ( y ). Therefore, f y = 2 x 3 + 3 xy 2 + c 0 ( y ). Setting this equal to N , we have 2 x 3 + 3 xy 2 + c 0 ( y ) = 2 x 3 + 3 xy 2 + 2 so c 0 ( y ) = 2 and c = 2 y . So f = 2 x 3 y + xy 3 + x + 2 y (+constant) . c) C starts at (1 , 0) and ends at ( - e π , 0), so Z C ~ F · d~ r = f ( - e π , 0) - f (1 , 0) = - e - π - 1. 4. Z C yx 3 dx + y 2 dy = Z 1 0 x 2 x 3 dx + ( x 2 ) 2 (2 xdx ) = Z 1 0 3 x 5 dx = 1 / 2 . 5. a) fl fl fl fl u x u y v x v y fl fl fl fl = fl fl fl fl 2 x/y - x 2 /y 2 y x fl fl fl fl = 3 x 2 /y . Therefore, dudv = (3 x 2 /y ) dxdy = 3 u dxdy = dxdy = 1 3 u dudv. b) Z 4 2 Z 5 1 1 3 u dudv = Z 4 2 1 3 ln 5 dv = 2 3 ln 5 . 6. a) I C Mdx = ZZ R - M y dA. b) We want M such that - M y = ( x + y ) 2 . Use M = - 1
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Unformatted text preview: = 2 y , so Z Z R 2 y dA = Z 1 Z x 3 2 y dydx = Z 1 x 6 dx = 1 7 . b) For the ±ux through C 1 , ˆ n =-ˆ implies ~ F · ˆ n =-(1 + y 2 ) =-1 where y = 0. The length of C 1 is 1, so the total ±ux through C 1 is-1. The ±ux through C 2 is zero because ˆ n = ˆ ı and ~ F ⊥ ˆ ı . c) Z C 3 ~ F · ˆ n ds = Z Z R div ~ FdA-Z C 1 ~ F · ˆ n ds-Z C 2 ~ F · ˆ n ds = 1 7-(-1)-0 = 8 7 ....
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