= Solution
For a <uniform random point in a disk>, the event $R<x$ occupies a <disk> of area $\pi x^2$ when $0<x<1$. Thus
$$
\boxed{F_R(x)=\begin{cases}0,&x\le0,\\x^2,&0<x<1,\\1,&x\ge1.\end{cases}}
$$
The radial <probability density function> is $f_R(r)=2r$ on $(0,1)$. Integrating gives
$$
\boxed{\mathbb ER=\int_0^1 2r^2\,dr=\frac23,\qquad
\mathbb ER^2=\frac12,\qquad\operatorname{Var}R=\frac1{18}.}
$$
Choose the polar angle $\Theta\in[0,2\pi)$. The Cartesian density is $1/\pi$ in the <disk>, and the <polar coordinates> <Jacobian determinant> is $r$. Hence
$$
f_{R,\Theta}(r,\theta)=\frac r\pi=(2r)\frac1{2\pi}\quad(0<r<1,\ 0\le\theta<2\pi).
$$
The product support and product density prove that \b[$R$ and $\Theta$ are <independent>], with uniform angle.
Reflection symmetry gives $\mathbb EX=\mathbb EY=0$ and $\mathbb E(XY)=0$, so \b[$\operatorname{Cov}(X,Y)=0$]. They are nevertheless not <independent>. The events $X>3/4$ and $Y>3/4$ each have positive <probability>, but their intersection is empty inside the <disk>, because it would require $X^2+Y^2>18/16>1$. <Independence> would give a positive product <probability> for that intersection.
Except at the probability-zero center, $X/R=\cos\Theta$ and $Y/R=\sin\Theta$. Therefore
$$
\boxed{\mathbb E\frac XR+i\mathbb E\frac YR
=\mathbb E e^{i\Theta}=\frac1{2\pi}\int_0^{2\pi}e^{i\theta}\,d\theta=0,
\qquad\xi=\Theta\pmod{2\pi}.}
$$
An arbitrary definition at the center does not change these <expected values>. This example distinguishes uncorrelated Cartesian coordinates from <independent> <polar coordinates>.
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