How to show that polar coordinates in a uniform distribution on a disk are independent? Let the random point $(X,Y)$ be uniformly distributed on the unit disc $D=\{(x,y):x^{2}+y^{2}<1\}$.
Show that the polar coordinates $R\in [0,1)$ and $\theta \in [0,2\pi)$ of the point are independent.
Can you help me with this exercise please?
 A: The solution may be a bit quirky, with a lot of variables, but it works fine for me.
We know that $X$,$Y$ - random variables in $\mathbb{R}^n$ and $\mathbb{R}^m$ are independent iff
$$\mathbb{E}(\varphi (X) \psi (Y) )= \mathbb{E}(\varphi(X)) \cdot \mathbb{E}(\psi(Y)) $$
$ \forall \varphi \in C^{\infty}_{0} $, $ \forall \psi \in C^{\infty}_{0} $, where $C^{\infty}_{0} $ are continuos on compact smooth functions.
If $(X,Y)$ are uniformly distributed on the unit disc $D=\{(x,y):x^{2}+y^{2}<1\}$. Then we have that $$\mathbb{P}_{(X,Y)}(dxdy)=\frac{1}{\pi}\mathbb{1}_D (x,y) dxdy$$
Analogically we define two distance and angle functions: $$\begin{cases}r:\mathbb{R^2}\rightarrow \mathbb{R}\\ \vartheta:\mathbb{R^2}\rightarrow \mathbb{R}\end{cases}$$
Now we have $R=r(X,Y)$ and $\Theta=\vartheta(X,Y)$ and by passing to polar coordinates with $\begin{cases}\rho=r(x,y)\\ \gamma = \vartheta(x,y) \end{cases}$ we can show that
$$\mathbb{E}(\varphi(R))=2 \int\limits^1_0 \varphi(\rho)\rho d\rho$$
$$\mathbb{E}(\psi(\Theta))=\frac{1}{2\pi}\int\limits^{2\pi}_{0} \psi (\gamma) d\gamma$$
Now we see that 
$$\mathbb{E}(\varphi(R)\psi(\Theta))=\frac{1}{\pi} \int \limits_{D} \varphi(r(x,y))\cdot \psi(\vartheta(x,y))dxdy=\\
\frac{1}{\pi}\int\limits^1_0[\int\limits^{2\pi}_{0} \varphi(\rho) \psi(\gamma)d\gamma]\rho d\rho=\frac{1}{\pi} \int\limits^1_0 \varphi(\rho) \rho d\rho \cdot \int\limits^{2\pi}_{0} \psi(\gamma)d\gamma= \\  \mathbb{E}(\varphi(R)) \cdot \mathbb{E}(\psi(\Theta))
$$
Using Fubini's theorem we prove that $R$ and $\Theta$ are independent.
