Assignment 9 Solutions

# Assignment 9 Solutions - Math 237 Assignment 9 Solutions 1...

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Unformatted text preview: Math 237 Assignment 9 Solutions 1. Use T ( x,y ) = ( x + y,- x + y ) to evaluate R π R π- 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 ≤ 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 6 = 0 . Hence, since the transformation has continuous partial derivatives we get by the change of variables theorem Z π Z π- y ( x + y ) cos( x- y ) dx dy = Z π Z u- u u cos(- v ) 1 2 dv du = 1 2 Z π- u sin(- v ) u- u du = 1 2 Z π 2 u sin u du =- u cos u + sin u π = 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 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 π, since RR D uv 1 dA just calculates the area of circle in the uv-plane. 2 3. Evaluate RRR R x 2 + y dV where R is the region bounded by...
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Assignment 9 Solutions - Math 237 Assignment 9 Solutions 1...

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