Evaluate
$$I=\iint\limits_D(x^2-2xy+y^2)\,\mathrm{d}x\,\mathrm{d}y$$
where $D$ is bounded by $$x^2-y^2=4,~x^2-y^2=16,~y=\frac{x}{2}-5,~y=\frac{x}{2}+1.$$
Attempt. The obvious transformation $u=x^2-y^2,~v=2y-x$ has Jacobian determinant
$$det\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{det\frac{\partial(u,v)}{\partial(x,y)}}=\frac{1}{4x-2y},$$ where $x=x(u,v)$ and $y=y(u,v)$. So: $$I=\frac{1}{2}\iint\limits_{[4,16]\times[-10,2]}\frac{x^2-2xy+y^2}{|4x-2y|}\,\mathrm{d}u\,\mathrm{d}v,$$ but the function inside the integral seems compicated enough, in order to turn into a simple function of $u,\,v.$
Thanks for the help.
The problem with $u = x^2 - y^2, v = 2y-x$ is that it is difficult to express the objective as a function of $u,v$
Which gets me thinking about
$u = x+y\\ v = x-y\\ du dv = 2 dx dy$
This works nice with the objective and one of the functions on the boundary, but not the other.
Our transformed objective is now $\frac 12 v^2$
and the boundaries are $uv = 4, uv = 16, u - 3v = -20, u-3v = 4$
Suppose we try:
$s = uv\\ t = u\\ v = \frac {s}{t} = v\\ u = t$
$du dv = \left|\begin{array}{} \frac{1}{t} &-\frac{s}{t^2}\\1 & 0\end{array}\right| = \frac {s}{t^2} ds dt$
After this transformation.
objective $\frac {s^3}{2}$
and the boundaries are $s = 4, s = 16, \frac{s}{t} - 3t = -20, \frac {s}{t}-3t = 4$
Can we express isolate $t$ in the following:
$ \frac {s}{t} - 3 t = -20\\ 3t^2 - 20 t - s\\ t = \frac {10 \pm \sqrt {100 + 3s}}{3}$
and
$t = \frac {2 \pm \sqrt {4 + 3s}}{3}$
At this point it is worth noting that if we sketch the region as it was originally defined there are, in fact two regions.
$\int_4^{16}\int_{\frac{2 - \sqrt{4+3s}}{3}}^{\frac{10 - \sqrt{100+3s}}{3}} s^3 \ dt\ ds + \int_4^{16}\int_{\frac{2 + \sqrt{4+3s}}{3}}^{\frac{10 + \sqrt{100+3s}}{3}} s^3 \ dt\ ds$
Rather than the two step transformation we could have said
$u = (x-y)(x+y)\\ v = x+y$
at the start. But, I didn't know that is what we needed.