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# A9_soln - Math 237 Assignment 9 Solutions 0 y(x 0 1 Use T(x...

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Math 237 Assignment 9 Solutions 1. Use T ( x, y ) = ( x + y, - x + y ) to evaluate π 0 π - y 0 ( x + y ) cos( x - y ) dx dy . Solution: We have u = x + y and v = - x + y . 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 . Under the mapping T we get: LINE 1: x = 0, 0 y π gives v = y = u with 0 u π . LINE 2: y = 0, 0 x π gives v = - x = - u with - π u 0. LINE 3: x + y = π , 0 x π gives u = π . We have u - v = 2 x , hence v = u - 2 x = π - 2 x so - π v π . Drawing rectangle vertically in the region we get - u v u and 0 u π . The Jacobian is ( x, y ) ( u, v ) = 1 2 - 1 2 1 2 1 2 = 1 2 = 0 . Hence, since the transformation has continuous partial derivatives we get by the change of variables theorem π 0 π - y 0 ( x + y ) cos( x - y ) dx dy = π 0 u - u u cos( - v ) 1 2 dv du = 1 2 π 0 - u sin( - v ) u - u du = 1 2 π 0 2 u sin u du = - u cos u + sin u π 0 = π 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 π . 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 0 1 = 1 = 0 . Hence since the mapping has continuous partial derivatives the change of variables theorem gives Area = D 1 dA = D uv 1 dA = 4 π, since D uv 1 dA just calculates the area of circle in the uv -plane.

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2 3. Evaluate R x 2 + y dV where R is the region bounded by x + y + z = 2, z = 2, x = 1 and y = x . Solution: Integrating wrt to z first gives 2 - x - y z 2 and then the region D xy is bounded by x = 1, y = x and the intersection of x + y + z = 2 and z = 2 so x + y = 0.
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