Integrals over a transformed region

[mcafej]

Homework Statement

Consider the change of variables x = x(u, v) = uv and y = y(u, v) =u^3+v^3

Compute the area of the part of the x-y plane that is the transform of the unit square in the
2nd quadrant of the u-v plane, which has one corner at the origin. (Since the transformation
is 1:1 on the second quadrant (assignment 6), the area equals the integral over the square of
the absolute value of the determinant of the Jacobian of the transformation.)

The Attempt at a Solution

So I computed the Jacobian to be 3v^3-3u^3. Then, since I just needed to integrate over a square, I did
∫$^{1}_{0}$∫$^{0}_{-1}$ 3v$^{3}$-3u$^{3}$ du dv. I keep getting 0 as an answer, but that just doesn't seem right, am I misunderstanding the question?

also, sorry if my formatting is confusing, I don't really know how to make the integrals look pretty

 

[haruspex]

∫$^{1}_{0}$∫$^{0}_{-1}$ 3v$^{3}$-3u$^{3}$ du dv. I keep getting 0 as an answer

I don't get zero. Try writing out the steps in more detail. I think you're getting a sign wrong.

 

[mcafej]

Wow...yea, i got a sign mixed up and ended up with .75-.75 instead of .75+.75. I got 1.5 as an answer, which sounds much more reasonable, thanks.