PDF of Sum of Two Random Variables $X$ and $Y$ are uniformly distributed on the unit disk. Thus,
$f_{X,Y}(x,y) = \begin{cases} \frac{1}{\pi}, & \text{if} ~ x^2+y^2 \leq 1,\\
0, &\text{otherwise.}\end{cases}$
If $Z=X+Y$, find the pdf of $Z$.

Following Dilip's hint, the geometry of the circular segment gives 
$$
F_Z(z) = \begin{cases} 0 & \text{if} \quad -\infty< z\leq -\sqrt{2} \\
\frac{1}{\pi} \left(\cos^{-1}\left(-\frac{z\sqrt{2}}{2}\right) + \frac{z\sqrt{2}}{2} \sqrt{1-\frac{z^2}{2}}\right) & \text{if} \quad-\sqrt{2}<z\leq 0 \\
1 - \frac{1}{\pi} \left(\cos^{-1}\left(\frac{z\sqrt{2}}{2}\right) - \frac{z\sqrt{2}}{2} \sqrt{1-\frac{z^2}{2}}\right) & \text{if} \quad 0<z<\sqrt{2} \\
1 & \text{if} \quad \sqrt{2}\leq z<\infty\quad  
\end{cases}
$$
which can be checked by this Monte Carlo simulation coded in R.
N <- 10^5
u <- runif(N, min = -1, max = 1)
v <- runif(N, min = -1, max = 1)
inside <- (u^2 + v^2 <= 1)
x <- u[inside]
y <- v[inside]
accepted <- sum(inside)
plot(x, y, xlim = c(-1, 1), ylim = c(-1, 1), pch = ".")
z = -0.9
sum(x + y <= z) / accepted
(acos(-z*sqrt(2)/2) + (z*sqrt(2)/2) * sqrt(1 - z^2/2)) / pi
z = 1.2
sum(x + y <= z) / accepted
1 - (acos(z*sqrt(2)/2) - (z*sqrt(2)/2) * sqrt(1 - z^2/2)) / pi

P.S. As Dilip's comment below show, it is much easier to work directly with the pdfs, because 
$$
  f_Z(z)=\frac{\sqrt{2}}{2}f_X\left(\frac{z\sqrt{2}}{2}\right) \, ,
$$
and 
$$
  f_X(x)=\frac{2}{\pi}\sqrt{1-x^2}\;I_{[-1,1]}(x) \, .
$$
To adjust the indicator, notice that $-1\leq z\sqrt{2}/2\leq 1$ if and only if $-\sqrt{2}\leq z\leq \sqrt{2}$. Hence, $I_{[-1,1]}(z\sqrt{2}/2)=I_{[-\sqrt{2},\sqrt{2}]}(z)$. 
