Finding the appropriate transformation to apply to an integral

[Pr0grammer]

Homework Statement

Evaluate $$\int_0^1 \! \int_0^{1-x} \! \sqrt{x+y} \left(y-2x\right)^{2} \, \, \mathrm{d}y \, \mathrm{d}x.$$ by applying the appropriate transformation.

Homework Equations

N/A

The Attempt at a Solution

So far, the best I can come up with is u=1-x, v=x+y, which gives me $$\int_0^1 \! \int_0^u \! \sqrt{v} \left(v-3-3u\right)^{2} \, \, \mathrm{d}v \, \mathrm{d}u.$$.

I know how to evaluate it after applying the transformation. As far as I can tell, one of the transformations should be 1-x to deal with the limit, and the other should be x+y to deal with the square root. I was wondering if there was a better way to do the transformation, though, since it still seems like it might be somewhat painful to integrate.

 

[micromass]

That was actually a very good transformation. The integral you end up with is not that hard, is it?

 

[Pr0grammer]

Wolfram suggests that the correct answer should be 2/9: http://www.wolframalpha.com/input/?i=int+(sqrt(x%2By)(y-2x)^2)+dy+dx,+x%3D0+to+1,+y%3D0+to+1-x+

However, after evaluating my transformation, I got 1892/315, which Wolfram agreed with: http://www.wolframalpha.com/input/?i=int+(sqrt(v)(v-3u-3)^2)+dv+du,+v%3D0+to+u,+u%3D0+to+1

Any idea where the mistake might be? (Also, I got -1 for the Jacobian from x=1-u and y=u+v-1. Is that right?)