$\iint (y^2-x^2) e^{xy} dxdy$ $\iint_G(y^2-x^2)e^{xy}dxdy$ on G = $\{ (x,y) | x \ge 0, 1 \le xy \le 4, 0 \le y-x \le 3 \}$
I have tried a substitution of 
$u = xy$
$v = y-x$
which got me to $\bar G = \{ (u,v) | 1 \le u \le 4, 0 \le v \le 3 \}$
and
$\iint_\bar G \left(\frac{v}{\sqrt{v^2+4u}}\right)^{\frac12} e^u dudv$
(note that $x = -\frac12v+\frac12\sqrt{v^2+4u}$
and that $y = \frac12v+\frac12\sqrt{v^2+4u}$ ) 
But from here on i'm stuck.
 A: Hint: As suggested in the comments, try substituting
$$\begin{cases}
&y-x=u\\
&x+y=v\\
\end{cases}$$
$$\implies\begin{cases}
&x=\frac{v-u}{2}\\
&y=\frac{u+v}{2}.\\
\end{cases}$$
The determinant of the Jacobian matrix for this linear transformation is found to be
$$\det{\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}}=-2$$
so then the differential areas are related as
$$\left|\det{\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}}\right|\mathrm{d}x\mathrm{d}y=\mathrm{d}u\mathrm{d}v\implies\mathrm{d}x\mathrm{d}y=\frac12\,\mathrm{d}u\mathrm{d}v.$$
Note then that $xy=\frac{v-u}{2}\cdot\frac{v+u}{2}=\frac{v^{2}-u^{2}}{4}$.
The planar region $G$ may be given in terms of $\left(u,v\right)$-coordinates as follows:
$$\begin{align}
G
&=\{\left(x,y\right)|0\le x\land1\le xy\le4\land0\le y-x\le3\}\\
&=\{\left(u,v\right)|0\le\frac{v-u}{2}\land1\le\frac{v-u}{2}\cdot\frac{v+u}{2}\le4\land0\le u\le3\}\\
&=\{\left(u,v\right)|u\le v\land4\le v^{2}-u^{2}\le16\land0\le u\le3\}\\
&=\{\left(u,v\right)|0\le u\le3\land u\le v\land\sqrt{4+u^{2}}\le v\le\sqrt{16+u^{2}}\}\\
&=\{\left(u,v\right)|0\le u\le3\land\sqrt{4+u^{2}}\le v\le\sqrt{16+u^{2}}\}.\\
\end{align}$$
Then, using the change of variables theorem, we find
$$\begin{align}
I
&=\iint_{G}\left(y^{2}-x^{2}\right)e^{xy}\,\mathrm{d}x\mathrm{d}y\\
&=\iint_{G}\left(y-x\right)\left(y+x\right)\exp{\left(xy\right)}\,\mathrm{d}x\mathrm{d}y\\
&=\iint_{G}uv\exp{\left(\frac{v^{2}-u^{2}}{4}\right)}\,\frac12\,\mathrm{d}u\mathrm{d}v\\
&=\int_{0}^{3}\mathrm{d}u\int_{\sqrt{4+u^{2}}}^{\sqrt{16+u^{2}}}\mathrm{d}v\,\frac{uv}{2}\exp{\left(\frac{v^{2}-u^{2}}{4}\right)}.\\
\end{align}$$
The iterated integral obtained in the last line above should be easy enough for anyone taking a multivariable calculus class, so I leave the rest as an exercise for our fearless reader. :)
A: A slightly different change of variables which simplifies a lot the boundary conditions :
$\begin{cases}
    s=y-x\\
    p=xy\\
  \end{cases}$
The Jacobian is :  $-\begin{vmatrix}
  -1 & 1 \\
  y & x \\ 
 \end{vmatrix}^{-1}=\frac{1}{y+x}$
$(y^2-x^2)\frac{1}{y+x}=y-x=s$
$\iint_{1 \le xy \le 4, 0 \le y-x \le 3}(y^2-x^2)e^{xy}dxdy =
\int_1^4\int_0^3 s e^p ds\;dp=
\int_1^4 e^p dp\int_0^3 s ds=
(e^4-e)\frac{3^2}{2}$ 
