Cdf of the joint density $f(x, y) = \frac{3}{2 \pi} \sqrt{1-x^2-y^2}$ $$f(x, y) = \frac{3}{2\pi} \sqrt{1-x^2-y^2}, \quad x^2 + y^2 \leq 1$$

Find the cdf $F(x, y)$. To do this, we need to compute the integral
$$ \int_{-1}^{x} \int_{-1}^{y} \frac{3}{2\pi} \sqrt{1-u^2-v^2} dv du .$$
This is where I'm stuck. Converting to polar coordinates wouldn't seem to help since the area to be integrated isn't a circle.
Edit: Adjusting the limits of integration to 
$$\int_{-1}^{x} \int_{-\sqrt{1-x^2}}^{y} \frac{3}{2\pi} \sqrt{1-u^2-v^2} dv du,$$
I tried using the substitution 
$$v=\sqrt{1−u^2}\sin{\theta}, \quad dv=\sqrt{1−u^2}cos{\theta}d\theta$$ 
Then $\theta = \arcsin(\frac{v}{\sqrt{1-u^2}})$, giving
$$\int_{-1}^{x} \int_{\arcsin(-\frac{\sqrt{1-x^2}}{\sqrt{1-u^2}})}^{\arcsin(\frac{y}{\sqrt{1-u^2}})} \frac{3}{2\pi} \sqrt{1-u^2-(\sqrt{1−u^2}\sin{\theta})^2} \sqrt{1−u^2}cos{\theta}d\theta du$$
$$=\int_{-1}^{x} \frac{3}{2\pi} (1−u^2) \int_{\arcsin(-\frac{\sqrt{1-x^2}}{\sqrt{1-u^2}})}^{\arcsin(\frac{y}{\sqrt{1-u^2}})} cos^2{\theta}d\theta du$$
$$=\int_{-1}^{x} \frac{3}{2\pi} (1−u^2) \int_{\arcsin(-\frac{\sqrt{1-x^2}}{\sqrt{1-u^2}})}^{\arcsin(\frac{y}{\sqrt{1-u^2}})} \frac{1}{2}[1+cos(2\theta)]d\theta du$$
$$=\int_{-1}^{x} \frac{3}{4\pi} (1−u^2) [\theta+\frac{1}{2}\sin(2\theta)]_{\arcsin(-\frac{\sqrt{1-x^2}}{\sqrt{1-u^2}})}^{\arcsin(\frac{y}{\sqrt{1-u^2}})} du$$
At this point the integral got really messy to deal with. Is there another way to approach this?

This problem will be easier to solve if you first try and visualize what the joint pdf looks as a surface above the $x$-$y$ plane in three-dimensional space. Hint: ignoring the scale factor $\frac{3}{2\pi}$, what is the surface defined by 
$$z = \begin{cases}\sqrt{1 - x^2 - y^2}, & x^2+y^2 \leq 1,\\
0, &\text{otherwise}, \end{cases}~~~~?$$ Can you think of ways that you might be able to compute the volume between this surface and the $x$-$y$ plane in the region $\{(x,y) \colon x\leq a, y \leq b\}$. The visualization will also help you set the lower and upper limits on the integrals as suggested by JarleTufto.
