= Solution
The <joint probability density> of the two <independent> <standard normal random variables> is $(2\pi)^{-1}e^{-(x^2+y^2)/2}$. The <polar coordinates> map has absolute <Jacobian determinant> $r$, so the <change of variables for probability densities> gives
$$
\boxed{f_{R,\Theta}(r,\theta)=\frac r{2\pi}e^{-r^2/2}
\mathbf1_{\{r>0,\ 0\leq\theta<2\pi\}}.}
$$
At the origin the angle may be assigned arbitrarily: that point has <probability> zero. The <joint probability density> factorizes as a <Rayleigh distribution> density $r e^{-r^2/2}$ and a <continuous uniform distribution> density $1/(2\pi)$. Hence $R$ and $\Theta$ are <independent>.
For $u>0$, substitute $s=r^2/2$ to find
$$
\mathbb P(R^2\leq u)=\int_0^{\sqrt u}r e^{-r^2/2}\,dr
=\boxed{1-e^{-u/2}.}
$$
Therefore $R^2\sim\operatorname{Exp}(1/2)$, equivalently $\chi_2^2$, and applying a measurable function to $R$ preserves its <independence> from $\Theta$. Explicitly, with $S=R^2$,
$$
\boxed{f_{\Theta,S}(\theta,s)=\frac1{4\pi}e^{-s/2}
\mathbf1_{\{0\leq\theta<2\pi,\ s>0\}}.}
$$
Thus $\Theta\sim\operatorname{Unif}[0,2\pi)$ and $S\sim\operatorname{Exp}(1/2)$ independently. This <polar decomposition of a planar standard normal vector> separates the uniformly distributed direction from its random squared distance.
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