Compute a $2$D integral $$I=\iint_{D}\cfrac{dxdy}{\sqrt{1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}}\times\left( {x}^{2}+{y}^{2}+1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}\right)^{\cfrac{3}{2}}}$$
where $$D=\left\{(x,y);\cfrac{{x}^{2}}{{a}^{2}}+\cfrac{{y}^{2}}{{b}^{2}}\leq 1,x\geq 0,y\geq 0\right\}.$$
How to compute then? What I can image is just the radial transformation: $x=ar\cos t, y=br\sin  t$.

Let 
$$\Delta(\theta) = a^2 \cos^2(\theta) + b^2\sin^2(\theta)$$
In terms of the parametrization $(x,y) = (ar\cos\theta,br\sin\theta)$, 
the integral $\mathscr{I}$ we want can be rewritten as
$$\begin{align}
\mathscr{I} =  & ab \int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{rdr}{\sqrt{1-r^2}\sqrt{\Delta(\theta) r^2 + 1 - r^2}^3}\\
= & \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{dr^2}{(1-r^2)^2}\frac{1}{\sqrt{\Delta(\theta) \frac{r^2}{1-r^2} + 1}^3}
\end{align}$$
Let $u = \frac{r^2}{1-r^2}$, $v = \Delta(\theta) u + 1$ and $t = \tan\theta$.
Notice $\displaystyle du = \frac{dr^2}{(1-r^2)^2} $, we have
$$\mathscr{I} = \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^{\infty} \frac{du}{\sqrt{\Delta(\theta) u + 1}^3}
= \frac{ab}{2}\int_0^{\frac{\pi}{2}} \frac{d\theta}{\Delta(\theta)} \int_1^{\infty} \frac{dv}{\sqrt{v}^3}\\
= ab \int_0^{\frac{\pi}{2}} \frac{d\theta}{a^2\cos^2(\theta) + b^2\sin^2(\theta)}
= ab \int_0^{\infty} \frac{dt}{a^2 + b^2 t^2}
= \frac{\pi}{2}
$$
