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?

$$
\mbox{With}\quad x \equiv u + v\quad\mbox{and}\quad y \equiv u - v
\quad\Longrightarrow\quad
{\partial\left(x,y\right) \over \partial\left(u,v\right)} = -2
$$
\begin{align}
&
\\[6mm]?&
=
\left.\vphantom{\Huge A}
\int_{0}^{\infty}{\rm d}x\int_{0}^{\infty}{\rm d}y\,\left(x^{2} - y^{2}\right)
\right\vert_{3\ \geq\ y - x\ >\ 0 \atop \vphantom{\Large A}5\ \geq\ xy\ >\ 2}
=
-2\left.\vphantom{\Huge A}
\int_{-\infty}^{\infty}{\rm d}v\int_{-\infty}^{\infty}{\rm d}u\,4uv
\right\vert_{3\ \geq\ -2v\ >\ 0 \atop {\vphantom{\Large A^{A}}5\ \geq\ u^{2}\ -\ v^{2}\ >\ 2\atop u\ >\ \left\vert v\right\vert}}
\\[4mm]&=
-8\int_{-3/2}^{0}{\rm d}v\ v\int_{\sqrt{v^{2} + 2}}^{\sqrt{v^{2} + 5}}{\rm d}u\ u
=
-8\int_{-3/2}^{0}{\rm d}v\ v\left({v^{2} + 5 \over 2} - {v^{2} + 2 \over 2}\right)
=
-12\,\left\lbrack -\,{\left(-3/2\right)^{2} \over 2}\right\rbrack
\\[4mm]&={\Large {27 \over 2}}
\end{align}
