<s>[INST] 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?
 [/INST] 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.
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