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FINAL_ChangeVar

# FINAL_ChangeVar - Z 1 Z 1 dxdy √ 1 x 2 y using x,y = T...

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Problem: Let D be the parallelogram with vertices (0 , 0), ( - 2 , 2), (2 , - 1) and (0 , 1). Evaluate ZZ D dxdy 1 + x + 2 y , using the transformation ( x, y ) = T ( u, v ) = ( - u + v, u - v 2 ). Solution: ZZ D f ( x, y ) dxdy = ZZ D * f ( x ( u, v ) , y ( u, v )) ( x, y ) ( u, v ) dudv, where ( x, y ) ( u, v ) = ∂x ∂u ∂x ∂v ∂y ∂u ∂y ∂v . Since x = - u + v and y = u - 1 2 v , f ( x ( u, v ) , y ( u, v )) = 1 p 1 - u + v + 2( u - v 2 ) = 1 1 + u and ( x, y ) ( u, v ) = - 1 1 1 - 1 2 = 1 2 - 1 = - 1 2 . Notice that T is a linear, 1-1 transformation. To get the vertices of the region D * you will need to find the inverse of T , i.e, solve for u and v in terms of x and y . We have x = - u + v (1) y = u - 1 2 v (2) Equation (1) plus equation (2) gives us v = 2 x + 2 y , similarly 1 2 eq (1) + eq (2) gives us u = x + 2 y . So ( u, v ) = ( x + 2 y, 2 x + 2 y ). And we get ( x, y ) ( u, v ) (0 , 0) (0 , 0) ( - 2 , 2) (2 , 0) (2 , - 1) (0 , 2) (0 , 1) (2 , 2) 1

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ZZ D dxdy 1 + x + 2 y = ZZ D * 1 1 + u - 1 2 dudv = 1 2 Z 2 0 Z 2 0 dudv 1 + u = 1 2 Z 2 0 2(1 + u ) 1 / 2 2 0 dv = Z 2 1 ( 3 - 1) dv = ( 3 - 1) v 2 0 = 2( 3 - 1) . Note for 3pm section: The problem posed in class was to evaluate
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Unformatted text preview: Z 1 Z 1 dxdy √ 1 + x + 2 y , using ( x,y ) = T ( u,v ) = (-u + v,u-v 2 ). In this case we get ( x,y ) ( u,v ) (0 , 0) (0 , 0) (1 , 0) (1 , 2) (1 , 1) (3 , 4) (0 , 1) (2 , 2) Integrating over D * is not simple. You will have to split and integrate over two regions: 2 So you get ZZ D dxdy √ 1 + x + 2 y = ZZ D * 1 √ 1 + u ± ± ±-1 2 ± ± ± dudv = ZZ D * 1 1 √ 1 + u ± ± ±-1 2 ± ± ± dudv + ZZ D * 2 1 √ 1 + u ± ± ±-1 2 ± ± ± dudv = 1 2 Z 2 Z v v/ 2 dudv √ 1 + u + 1 2 Z 4 2 Z ( v +2) / 2 v-1 dudv √ 1 + u = ... Clearly this given T is not a good transformation to use in this case!!! 3...
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FINAL_ChangeVar - Z 1 Z 1 dxdy √ 1 x 2 y using x,y = T...

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