I'm trying to find the sum of two independent variables $A$ and $B$ whose densities are defined as follows:
$$f_A(x,y)=\frac{1}{\pi}, \sqrt{x^2+y^2}\le1$$
and
$$f_B(x,y)=\frac{1}{\pi}, \sqrt{x^2+y^2}\le1$$
Basically, I need to find the PDF of $$f_{A+B}(x,y)$$. How can I do that? I understand that I need to convolve the two densities, and I guess at a certain step converting x and y to polar coordinates is inevitable, but I still have very little idea of how to actually proceed.
Let $S = A + B$ for $x, y\in \mathbb R$, such that $r = \sqrt{x^2 + y^2} < 2$
\begin{align} f_S(x, y) &= \frac1{\pi^2}\int_{\mathbb R^2} \mathbf 1_{t^2 + s^2 \le 1} \mathbf 1_{(x-t)^2 + (y-s)^2 \le 1}\mathrm ds\mathrm dt\\ &= \frac1{\pi^2}\int_{\mathbb R^2} \mathbf 1_{t^2 + s^2 \le 1} \mathbf 1_{\left(r-t\right)^2 + s^2 \le 1}\mathrm dt\mathrm ds\\ &= \frac2{\pi^2}\int_{\frac r2}^1 \int_{-\sqrt{1-s^2}}^{\sqrt{1-s^2}}\mathrm dt \mathrm ds\\ &= \frac{4}{\pi^2}\int_{\frac r2}^1 \sqrt{1-s^2}\mathrm ds\\ &= \frac{4}{\pi^2}\int_{\arcsin\left(\frac r2\right)}^{\frac\pi2} \cos^2\theta\mathrm d\theta\\ &= \frac4{\pi^2} \left[\frac\theta2 + \frac{\sin2\theta}{4}\right]_{\arcsin\left(\frac r2\right)}^{\frac\pi2}\\ &= \frac4{\pi^2} \left(\frac\pi4 - \frac12\arcsin\left(\frac r2\right) - \frac14 \sin\left(2\arcsin\left(\frac r2\right)\right)\right)\\ &= \frac1\pi - \frac2{\pi^2}\arcsin\left(\frac r2\right) - \frac r{\pi^2}\sqrt{1 - \frac{r^2}4} \end{align}