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 $Z$.
 A: Since my hint seems to have not been very useful to you, 
here is a step by step procedure.
It is different from things involving Jacobians and simulations using
R and such stuff that will appear in
other answers to your question.
The key idea is that we will find $F_Z(z)$,
the cumulative probability distribution
function (CDF) of $Z = X+Y$. The CDF is, of course, a probability:
specifically, for any real number $z$.
$$F_Z(z) = P\{Z \leq z\}.$$
Once we have found the CDF for all values of $z$,
we can take the derivative of the CDF
with respect to $z$ to find the
probability density function $f_Z(z)$. I will leave the computation of the
derivative to you.
The probability that $(X,Y)$ lies in any region $A$ of the plane is
the integral of the joint pdf of $X$ and $Y$ over the region $A \cap D$
where $D$ is the unit disc. In this instance, since the joint pdf has
constant value on $D$, the calculation is much simpler: we merely need
to find the area of the region $A \cap D$ and divide by the area of
$D$, which you know is just $\pi$.  The trick in this problem is to
understand that in some cases, it is not necessary to evaluate
an integral to find an area if we can find the area using mensuration
formulas such as $\frac 12 \times~$ base $\times~$ altitude from 
middle school or high school.
So, draw the line $x+y = 0$ (overlaying it on the unit disc)
and consider the question: what is 
the probability that $(X,Y)$ lies on or below this line? Note that we
are asking for $$P\{X+Y \leq 0\} = P\{Z \leq 0\} = F_Z(0).$$
Can you figure this out just by looking at the picture you have drawn?
If you cannot, set up a double integral and evaluate it: conversion
to polar coordinates will be helpful. After you get the answer
(it is same as the answer to your chances of winning the lottery based
on the idea that there are only two outcomes: win or lose
and so it is a 50-50 chance that you win), look at the picture again
and try and figure out a way to get to the end result without integrating.
Let's try a slightly harder one. Draw the line $x+y = 1$ and figure out
$$P\{X+Y \leq 1\} = P\{Z \leq 1\} = F_Z(1).$$
You might find it easier to look at the first quadrant separately
from the other three quadrants.
More generally, for any fixed number $z \in [0,\sqrt{2}]$, 


*

*draw the line $x+y=z$, 

*find the points $(a,b)$ and $(c,d)$ where the line
intersects the unit circle,

*draw radius vectors from $(0,0)$ to $(a,b)$ and $(c,d)$
This divides the region of interest into a sector of the
unit disc and a triangle whose areas you should be able to
compute using formulas such as $\frac 12 r^2\theta$ where 
$\theta$ is the angle subtended by an arc at the center etc.
Next, you need to do all this for $z \in [-\sqrt{2},0]$ but
if you look at your pictures, you will find, I hope, that
a lot of work is already done if you can think of symmetry
and the notion that $P(B^c) = 1-P(B)$.
Finally, you need to find $F_Z(z)$ for $z > \sqrt{2}$ and for
$z < -\sqrt{2}$ but these should be easy.  This gives
you $F_Z(z)$ for all $z$. Differentiate, and you are done.
A: 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, $I_{[-1,1]}(z\sqrt{2}/2)=I_{[-\sqrt{2},\sqrt{2}]}(z)$. 
