How to find the integral by changing the coordinates? Let R be the region in the first quadrant where
$$3 \geq y-x \geq 0$$
$$5 \geq xy \geq2$$
Compute
$$\int_A (x^2-y^2)\,dx\,dy.$$
I tried to use $ u= y-x, v= xy$ as my change of coordinates, but then I don't know how to solve it.
Can someone help me?

Hint:  try the tranformation
$$v:=x+y,~~u:=x-y$$
or $x=\frac{u+v}{2}$, $y=\frac{v-u}{2}$. Now the integrand becomes the product $uv$ (up to scalar coming from the determinant of the Jacobian transformation) and the domain of integration is
$$A=\{(u,v): -u\geq 0, -u\leq 3, v^2-u^2\geq 8, v^2-u^2\leq 20 \}=
\{(u,v):  -3\leq u\leq 0, v^2-u^2\geq 8, v^2-u^2\leq 20 \} $$
The integral can now be computed.
