Solving Change of Variables for Triangular Region

[bodensee9]

I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!

 

[HallsofIvy]

I am wondering if someone could help me with the following? I am supposed to show that ∫∫f(x+y)dA evaluated from the triangular region with the vertices (0,0), (1,0) and (0,1) is equal to ∫∫uf(u)du.

This triangular region has the equations, x = 0, x = 1, and y = -x + 1. If I set x+y = u, then I know that u must satisfy the equations u = 1, u = 0. But, I’m not sure what to do after that? If someone could give me some hints, that be great. Thanks!

No, the triangular region has boundaries x= 0, y= 0, and x+ y= 1. The upper line, x+ y= 1 corresponds to u= 1 but neither x=0 nor y= 0 corresponds to u= 0. In any case, your problem is incomplete. ∫∫f(x+y)dA is a double integral. In ∫∫uf(u)du du is not an area differential. Did you mean show that it is equal to ∫∫uf(u)dudx? If so then the limits of integration are irrelevant. You need only show that dA= dydx= ududx. for some choice of u.