Let the random point (X,Y) be uniformly distributed on the square Let the random point $(X,Y)$ be uniformly distributed on the square 
$D=\{(x,y):-1\leq x\leq 1,\ -1\leq y \leq 1\}$.
Find the distribution function and the probability distribution function of $Z=X +Y$
 A: Hint: convolution.
Maybe Macro is right and the Wikipedia entry could be a little clearer.  Let's work through a bit of the problem together.  We see that in the (axis-aligned) square, knowing $X$ tells you nothing about $Y$ (and vice versa).  That is, they are independent.  So, you are looking for the sum of independent variables.  This is called a convolution.
So, let's try to figure out how a convolution should work.  We know that if $Z=0.2$, then one possibility is $X=-0.3, Y=0.5$; another is $X=0.1, Y=0.1$, and so on.  So, it might a bit of a leap here, to see that the density of $Z$ is:
$$f_Z(z) \propto \int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx$$
In other words, the density of $Z$ at $z$ is proportional to to the density of $X$ at $x$ times the density of $Y$ at $y$ where $y$ is the number that would have to be added to $x$ to get $z$: $z-x$.
We said proportional to because we still need the density of $Z$ to sum to one.
To complete the problem, try to simplify this integral.  What is the density of $X$?
A: For any random variables $X$ and $Y$ (not necessarily restricted to the unit square, and not necessarily independent) with joint density function $f_{X,Y}(x,y)$, and any real number $z$
we have that
$$P\{X + Y \leq z\} = \int_{x=-\infty}^{\infty} \int_{y = -\infty}^{z-x}
f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx$$
which integral can be easily set up by beginning students if they would
only bother to sketch the $x$-$y$ plane, draw on it the line $z = x+y$,
and think a bit about which of the two regions (above the line or below
the line) corresponds to the event $\{X+Y\leq z\}$.  Thus with $Z$
denoting $X+Y$, we can express the cumulative probability distribution
function (CDF) of $Z$ as 
$$F_Z(z) = P\{Z \leq z\} = P\{X + Y \leq z\} 
= \int_{x=-\infty}^{\infty} \int_{y = -\infty}^{z-x}
f_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx$$
and so we can find the density function $f_Z(z)$ by differentiating
the CDF thus obtained. Those familiar with Leibniz's rule for
differentiating under the integral sign will be able to bypass this 
two-step process and write
$$f_Z(z) = \int_{-\infty}^{\infty} f_{X,Y}(x,z-x)\,\mathrm dx$$
directly. Note the lack of a proportionality sign as in
NeilG's answer: this is an equality.
For the special case of $(X,Y)$ taking on values in a square
region as in the OP's problem, the computation of $F_Z(z)$
becomes an integration of the joint density over a right 
triangular region or a pentagonal region that is the complement of a
right triangular region. For the even more special case of a uniform
density, even formal integration can be avoided since simple mensuration
formulas suffice. This latter is the geometric approach mentioned
by cardinal in his comments on NeilG's answer.
