Double integral over complicated region Suppose we wanted to compute $\iint\frac {1}{1 + x^2 + y^2} dxdy$ over the region $\frac {(x-1)^2}4 + \frac {(y+2)^2}9 \leqslant 1$. 
It gets quite hairy if we use elliptical polar coordinates i.e. $(x,y) = (1 + 2rcos \theta, -2 + 3r\sin \theta ) $, so say we insist on letting $x^2 + y^2 = r^2$ with $(x,y) = (rcos \theta, r \sin \theta)$ in order to make the integrand simple.
How can we then use our insisted change of coordinates $(x,y) = ( r \cos \theta, r \sin \theta ) $ to describe the region $\frac {(x-1)^2}4 + \frac {(y+2)^2}9 \leqslant 1$ and actually carry out the integration?
The example function chosen, $\frac {1}{1 + x^2 + y^2}$, may perhaps been a bad one but hopefully the point comes across.
 A: It is worth to calculate the integral regarding the ellipse as a normal domain, from which:
$$\begin{eqnarray*} I &=& \int_{-1}^{3}\int_{-2-3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}^{-2+3\sqrt{1-\left(\frac{x-1}{2}\right)^2}}\frac{1}{1+x^2+y^2}\,dy\,dx.\end{eqnarray*}$$
Now exploiting the fact that $\int\frac{dy}{1+x^2+y^2}=\frac{1}{\sqrt{1+x^2}}\arctan\left(\frac{y}{\sqrt{1+x^2}}\right)$ and the identity:
$$\arctan(a)+\arctan(b)=\arctan\left(\frac{a+b}{1-ab}\right)$$
we have:
$$I = \int_{-1}^{3}\frac{1}{\sqrt{1+x^2}}\arctan\left(\frac{12\sqrt{(1+x^2)(3-x)(x+1)}}{13x^2-18x-7}\right)\,dx$$
and by setting $x=\sinh t$ we arrive at:
$$ I = \int_{\log(\sqrt{2}-1)}^{\log(3+\sqrt{10})}\arctan\left(\frac{12\cosh t\sqrt{(3-\sinh t)(1+\sinh t)}}{13\sinh^2 t-18\sinh t-7}\right)dt$$
that is hard to evaluate in terms of elementary functions but quite easy to calculate numerically. As an alternative, by setting $x=1+2\rho\cos\theta,y=-2+3\rho\sin\theta$ we have:
$$ I = \int_{0}^{1}\int_{-\pi}^{\pi}\frac{\rho}{6+4\rho\cos\theta-12\rho\sin\theta+4\rho^2\cos^2\theta+9\sin^2\theta}\,d\theta\,d\rho$$
and by setting $\theta=2\arctan t$ we have:
$$ I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{2\rho(1+t^2)}{6(1+t^2)^2+4\rho(1-t^4)-24\rho t(1+t^2)+4\rho^2(1-t^2)^2+36t^2}\,dt\,d\rho,$$
$$ I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{2\rho(1+t^2)}{(6-4\rho+4\rho^2)t^4-24\rho t^3+(48-8\rho^2)t^2-24\rho t+(6+4\rho+4\rho^2)}\,dt\,d\rho,$$
$$ I = \int_{0}^{1}\int_{-\infty}^{+\infty}\frac{\rho(1+t^2)}{2(1-t^2)^2\rho^2+(2-12t-12t^3-2t^4)\rho+(3+24t^2+3t^4)}\,dt\,d\rho.$$
