= Solution
The unit disk has area $\pi$, so its joint density is $f_{X,Y}(x,y)=\pi^{-1}\mathbf1_{\{x^2+y^2\le1\}}$. The polar-coordinate Jacobian has absolute <determinant> $r$. Therefore
$$
f_{R,\Theta}(r,\theta)=\frac r\pi,\qquad 0<r<1,\quad0\le\theta<2\pi,
$$
and is zero outside this rectangle, up to irrelevant boundary choices. Integrating gives
$$
\boxed{f_R(r)=2r\mathbf1_{(0,1)}(r),\qquad f_\Theta(\theta)=\frac1{2\pi}\mathbf1_{[0,2\pi)}(\theta),\qquad\mathbb ER=2/3}.
$$
The joint density equals the product of the two marginal densities, so \b[$R$ and $\Theta$ are independent]. The angle at the centre can be chosen arbitrarily since that point has probability zero.
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