Probability density of semicircle Let $(X,Y)$ be uniformly distributed on the unit circle. Find the density of
$R=(X^2+Y^2)^{1/2}$
Is this $1/2$ the probability density of the entire unit circle?
I was thinking setting up $F_R(r)=P(R<r)=P((X^2+Y^2)^{1/2} \leq R)$ as an integral.
 A: First note that since $(X,Y)$ is uniformly distributed within the unit circle, we have
$$\mathbb{P}((X,Y) \in (x,x+dx) \times (y,y+dy)) = \dfrac{dxdy}{\pi}$$ assuming $(x,y)$ is inside the unit circle. Hence,
$$\mathbb{P}(R \leq r) = \mathbb{P}(X^2+Y^2 \leq r^2) = \dfrac{\pi r^2}{\pi} = r^2$$
A: Setting up that integral is a good way to find the CDF (cumulative density function), which you can then differentiate to find the PDF (probability density function).  
From there, we'd have
$$
F_R(r) = \iint_{\{\sqrt{x^2 +y^2}<r\}}P(x,y)\,dx\,dy
= \int_0^{2\pi}\int_{0}^{r} P(x(\rho,\theta),y(\rho,\theta))\rho\,d\rho\,d\theta
$$
A: If $(X,Y)$ is uniformly distributed within the unit disk, and $A$ is a region within the unit disk, then the probability that $(X,Y)$ is in $A$ is the ratio of the area of $A$ to the area of the unit disk. 
We find the probability that $R\le r$. For $0\lt r\lt 1$, this is the area of the disk of radius $r$, divided by the area of the unit circle. So it is $\frac{\pi r^2}{r}$, that is, $r^2$.
So if $F_R(r)$ is the cdf of $R$, we have $F_R(r)=r^2$ for $0\lt r\lt 1$.
For completeness, note that $F_R(r)=0$ if $r\le 0$, and $F_R(r)=1$ if $r\ge 1$.
For the density function of $R$, differentiate. We get $f_R(r)=0$ if $r\le 0$, $f_R(r)=2r$ if $0\lt r\lt 1$, and $f_R(r)=0$ if $r\gt 1$. 
