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Unformatted text preview: Math 237 Assignment 9 Solutions 1. 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 0 x  y and 0 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, 0 y gives v = y = u with 0 u . LINE 2: y = 0, 0 x gives v = x = u with u 0. LINE 3: x + y = , 0 x gives u = . We have u v = 2 x , hence v = u 2 x =  2 x so v . Drawing rectangle vertically in the region we get u v u and 0 u . The Jacobian is ( x,y ) ( u,v ) = 1 2 1 2 1 2 1 2 = 1 2 6 = 0 . Hence, since the transformation has continuous partial derivatives we get by the change of variables theorem Z Z  y ( x + y ) cos( x y ) dx dy = Z Z u u u cos( v ) 1 2 dv du = 1 2 Z  u sin( v ) u u du = 1 2 Z 2 u sin u du = u cos u + sin u = 2. Find a linear transformation that maps x 2 + 4 xy + 5 y 2 = 4 onto a unit circle. Hence show that the area enclosed by the ellipse equals 4 . Solution: Completing the square we get x 2 + 4 xy + 5 y 2 = 4 is ( x + 2 y ) 2 + y 2 = 4. Hence, we let u = x +2 y and v = y . Thus, the inverse transformation is x = u 2 v , y = v . Which maps the circle of radius 2 onto the ellipse. The Jacobian is ( x,y ) ( u,v ) = 1 2 1 = 1 6 = 0 . Hence since the mapping has continuous partial derivatives the change of variables theorem gives Area = ZZ D 1 dA = ZZ D uv 1 dA = 4 , since RR D uv 1 dA just calculates the area of circle in the uvplane. 2 3. Evaluate RRR R x 2 + y dV where R is the region bounded by...
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This note was uploaded on 01/04/2012 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.
 Spring '08
 WOLCZUK
 Math

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