Random vector $(X,Y)$ is uniformly distributed on the disk $D_r$ defined by $$D_r=\{(x,y)\in \mathbb R^2\mid x^2+y^2\leq r\}.$$ Find the joint distribution of $R=\sqrt{X^2+Y^2}$ and $\Theta=\arctan\frac{Y}{X}$.
Need a kick starter or any hints. Any help is appreciated.
The common cumulative distribution of $R=\sqrt{X^2+Y^2}$ and $\Theta=\arctan\left(\frac YX\right)$ is by definition
$$F_{R,\Theta}(p,t)=P(R<p\cap\Theta<t)=\frac{T_{red}}{r^2\pi}=\frac{p^2\frac{t}{2\pi}}{r^2}, \,\,\, 0\le p\le r,\,\,0\le t \le 2\pi$$ where $T_{red}$ is the area of the red region depicted below.
The common density can be calculated by taking the following second order partial derivative:
$$f_{R,\Theta}(p,t)=\frac{\partial^2}{\partial p\partial t}F_{R,\Theta}(p,t)=\begin{cases} \frac p{\pi r^2},& \text{ if } 0\le p\le r,\,\,0\le t \le 2\pi\\ 0,&\text{ otherwise. } \end{cases}$$