A10_soln - Math 237 Assignment 10 Solutions 1. Let D xy be...

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Unformatted text preview: Math 237 Assignment 10 Solutions 1. Let D xy be the region in the xy-plane enclosed by the lines y = 2- x , y = 4- x , y = x and y = 0 . Let ( x,y ) = F ( u,v ) = ( u + uv,u- uv ) . a) Show that F has an inverse map F- 1 de ned on D xy . Solution: The Jacobian of the given map is ( x,y ) ( u,v ) = 1 + v u 1- v- u =- 2 u. Observe that if u = 0 then ( x,y ) = (0 , 0) which is not in D xy . Thus since F has continuous partial derivatives by the inverse property of the Jacobian and the inverse mapping theorem we get that F- 1 exists at each point in D xy . b) Find the image D uv of D xy under F- 1 . Solution: Observe that x + y = 2 u and x- y = 2 uv , hence u = x + y 2 and v = x- y 2 u = x- y x + y . Mapping the boundary lines we get LINE 1: y = x , 1 x 2 gives x- y = 0 hence v = 0 x = u so 1 u 2 . LINE 2: y = 0 , 2 x 4 gives u = x 2 so 1 u 2 and v = 1 . LINE 3: y = 2- x , 1 x 2 implies x + y = 2 u = 1 and v = x- y 2 = x- (2- x ) 2 = x- 1 so v 1 . LINE 4: y = 4- x , 2 x 4 implies x + y = 4 u = 2 and v = x- y 4 = x- (4- x ) 4 = 1 2 x- 1 so v 1 . Sketching gives: c) Use the mapping F to evaluate ZZ D xy e x- y x + y x + y dx dy . 2 Solution: By the Change of Variable Theorem, ZZ D xy e x- y x + y x + y dx dy = Z 1 Z 2 1 e v 2 u |- 2 u | du dv = Z 1 e v dv = e- 1 . 2. Use T ( x,y ) = ( x + y,- x + y ) to evaluate R R - y ( x + y ) cos( x- y ) dx dy . Solution: We have u = x + y and v =- x + y . The region of integration is x - y and y . Thus, the region is bounded by the lines x = 0 , y = 0 and x = - y . Under the mapping T we get: LINE 1: x = 0 , y gives v = y = u with u . LINE 2: y = 0 , x gives v =- x =- u with u . LINE 3: x + y = , x gives u = . We have u- v = 2 x , hence v = u- 2 x = - 2 x so- v ....
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This note was uploaded on 11/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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A10_soln - Math 237 Assignment 10 Solutions 1. Let D xy be...

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