recommended_problem_soln

# recommended_problem_soln - Math 237 Recommended Problem...

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Math 237 Recommended Problem Solutions Problem Set 9 A1. Use T ( x, y ) = ( x + y, - x + y ) to evaluate R π 0 R π - y 0 ( x + y ) cos( x - y ) dx dy . Solution: We have u = x + y and v = - x + y . Adding these we get y = 1 2 ( u + v ) and subtracting we get x = 1 2 ( u - v ). The region of integration is 0 x π - y and 0 y π . Thus, the region is bounded by the lines x = 0, y = 0 and x = π - y . In the uv -plane these are 1 2 ( u - v ) = 0 v = u , 1 2 ( u + v ) = 0 v = - u , and x + y = π u = π . Moreover, the Jacobian is ( x, y ) ( u, v ) = ± ± ± ± 1 2 - 1 2 1 2 1 2 ± ± ± ± = 1 2 . Hence, since the transformation is one-to-one we get Z π 0 Z π - y 0 ( x + y ) cos( x - y ) dx dy = Z π 0 Z u - u 1 2 u cos( - v ) dv du = 1 2 Z π 0 - u sin( - v ) ± ± ± ± u - u du = 1 2 Z π 0 2 u sin u du = - u cos u + sin u ± ± ± ± π 0 = π A4. Let D be the region enclosed by the lines y = 2 - x , y = 4 - x , y = x and y = 0. Evaluate the Jacobian of F ( u, v ) = ( u + uv, u - uv ). Sketch the image of D under F . Use the map to evaluate RR D e x - y x + y x + y dx dy . Solution: The Jacobian of the given map is ( x,y ) ( u,v ) = ± ± ± ± 1 + v u 1 - v - u ± ± ± ± = - 2 u. If u were equal to 0, then ( x, y ) = (0 , 0) which is not in the region D . Hence the Jacobain is never zero on D . The images of the given lines under the transformation are:

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## This note was uploaded on 04/18/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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recommended_problem_soln - Math 237 Recommended Problem...

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