TITLE: Distribution of $R^2 = X^2 +Y^2$ where (X,Y) is a point on the unit circle
QUESTION [6 upvotes]: So I have a point $(X,Y)$ chosen from the unit disk with uniform distribution. And I'm attempting to find the distribution of $R^2$, where $R$ is the distance from the point to the origin.  
Now obviously the joint distribution:  
$f_{X,Y}(x,y)=\frac{1}{\pi}, \forall (x,y)\in C$
where $C=\{(x,y)\in\mathbb{R}:x^2+y^2\le1\}$ 
Now, I attempted to take the marginal distributions, square them, and then add them back, which gave the following:  
$F_X(x)=F_Y(x)=\frac{2}\pi\sqrt{1-x^2}, \forall x\in [-1,1]$
Applying the transformation, $g:[-1,1]\to\mathbb{R}, x\mapsto x^2$  
\begin{align}
f_{X^2}(x) & =f_{Y^2}(x)=f_X(\sqrt{x}) \left|\frac{1}{2}x^{-1/2}\right| +  
\space f_X(-\sqrt{x}) \left| -\frac{1}{2}x^{-1/2}\right| \\[10pt]
& =\frac{2}\pi\sqrt{\frac{1-x^2}{x}}, \forall x\in[0,1]
\end{align}
But then I realised I couldn't just add them as $X$ and $Y$ are dependent and as such so would $X^2$ and $Y^2$.  
So I assume I have to apply the transformation, $h:[-1,1]\times[-1,1] \to \mathbb{R}^2, (x,y)\mapsto(x^2,y^2)$
Which would entail me finding the Jacobian and inverse, etc. which I am fully able to do.
However, $h$ is clearly not 1:1, and although I could get around that in the single variable case, I'm not sure what the anologue is for multy variables.  
So if anyone has any ideas on where to go from here, any assistance would be appreciated.

REPLY [3 votes]: Hint: if the points are uniformly distributed on the unit disc then density measure for all points on a circle of radius ($r$) will be proportional to the circumference of that circle.
So $f_R(r) = cr\,\mathbf 1_{r\in[0;1]}$ and $\int_0^1 cr\operatorname d r=1$ and hence $c=2$
Hence $f_R(r)=2r \,\mathbf 1_{r\in[0;1]}$
Use a simple change of variables transformation to find $f_{R^2}$
$$f_{R^2}(\rho) = f_R(\surd\rho)\cdot\left\lvert\dfrac{\operatorname d \surd\rho}{\operatorname d \rho}\right\rvert$$