Caculate $\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$, with D:$\frac{x^2}{2}+y^2\leq1$ I found some difficulty with this exercise:
Calculate
$$\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
with $D := \left\{(x,y)\in\mathbb{R}^{2}\mid \dfrac{x^2}{2}+y^2\leq1\right\}$
I use change of Variables in Polar Coordinates, but the integral become so hard to calculate.
I think maybe we change variables $u = x^2 + y^2$, the integral will be easier, but I can't find $v(x,y$) to have the Jacobi easy to calculate.

Take $$I=\iint_D \frac{x^2+y^2}{\sqrt{4-(x^2+y^2)^2}}dxdy$$
Consider the change of variables $(x,y)\to(r\cos\theta,r\sin\theta)$.
Let $\partial D=\{(x,y)\in\mathbb{R}^2:\frac{x^2}2+y^2=1\}$. Then for all $(x,y)=(r\cos\theta,r\sin\theta)\in\partial D$, $r^2(1-\frac12\cos^2\theta)=1$. Hence,
$$I=\int_0^{2\pi}\left(\int_0^{R(\theta)}\frac{r^2}{\sqrt{4-r^4}}rdr\right)d\theta$$
where $R(\theta)=\frac1{\sqrt{1-\frac12\cos^2\theta}}$. Now,
$$I=\int_0^{2\pi}\frac12\left(2-\sqrt{4-[R(\theta)]^2}\right)d\theta$$
$$\implies I=2\pi-\frac12\int_0^{2\pi}\sqrt{4-\left(1-\frac12\cos^2\theta\right)^{-2}}d\theta$$
