$\iint_{\mathbb R^2}\sqrt{\frac{x^2}{a^2}+\frac{x^2}{b^2}}e^{-\frac{x^2}{a^2}+\frac{y^2}{b^2}}dxdy$ $$\iint_{\mathbb R^2}\sqrt{\frac{x^2}{a^2}+\frac{y^2}{b^2}}\,e^{-\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}\,dx\,dy$$
Basically I have done problems similar to this, using the theorem that if $f>0$ then if $$\bigcup_{n=1}^{\infty} D_n=D$$ then $$\exists\int_{D}f(x)dx \Leftrightarrow \exists \lim_{n\to \infty}\int_{D_n}f(x)dx$$
So basically if I had the integral $$\iiint_{x^2+y^2+z^2\geq1}\frac{e^{-x^2-y^2-z^2}}{\sqrt{x^2+y^2+z^2}}$$ I would use the spherical coordinates substitution and have in this particular case $D_n=\{(x,y,z)|1<x^2+y^2+z^2<n^2\}$ and the radius in the mapping $(r,\phi,\theta)\mapsto (r\cos\varphi \sin\theta, r\sin \varphi \sin \theta,r\cos\theta)$ the ranges of the angles are clear as day, and the range of $r$ is from $1$ to $n$. 
Here in the main problem that I give, it seems like a could Idea to have $D_n$ be rising ellipses  and the substitution be $x=a\cos\phi, y=b\sin \phi$. The problem I have is setting up how this would actually look, could $D_{a,b}=\{(x,y)|\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\}$? I would appreciate if someone could bring this to a finished point, because I am not sure how to solve this one.   

By setting $x=au$ and $y=bv$ the problem boils down to computing
$$ I(a,b) = ab\iint_{\mathbb{R}^2}\sqrt{u^2+v^2} e^{-(u^2+v^2)}\,du\,dv = 2\pi ab \int_{0}^{+\infty} \rho^2 e^{-\rho^2}\,d\rho = \pi a b\cdot\Gamma\left(\frac{1}{2}\right).$$
