CDF and PDF of radius of a unit disk Let X and Y be uniformly distributed on a unit disk such that 
$x^2 + y^2 \leq 1$
Let $R = \sqrt{X^2 + Y^2}$.  What are the CDF and PDF of $R$?
I know that the area of the unit disk is 
$A = \pi r^2 = \pi 1^2 = \pi$
Thus, I think that the joint PDF of $X$ and $Y$ is the following, but 
I am not sure about this:
$f_{X, Y}(x, y) = \frac{1}{\pi}, \ \ \ x^2 + y^2 \leq 1$
I know that 
$P(R \leq r) = P(\sqrt{X^2 + Y^2} \leq r)$.
This is where I'm stuck.
 A: You are extremely close to the answer.  You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$.  Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:
$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2
\quad \quad \quad 0 \leqslant r \leqslant 1.$$
The reasoning behind this is that with a uniform distribution we have a flat density over the circle.  So the probability of falling within a given area is the relative size of that area compared to the size of the circle.  This gives the corresponding density function:
$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$
As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ from the origin to the point does not have a uniform distribution; as expected, it is more likely to be nearer to the outer edge of the circle than its origin.
