 Solution

A marginal density is obtained by integrating out the other variable: $\boxed{f_X(x)={\int }_{\mathbb{R}}{f}_{X,Y}(x,y)dy}$. Here independence gives $f_{X,Y}=e^{-(x^2+y^2)/2}/(2\pi )$.

The polar transformation has absolute Jacobian determinant $r$. Consequentlywith density zero outside this range. Integrating over the angle gives $\boxed{f_R(r)=r{e}^{-r^2/2}}$, the Rayleigh distribution density. Integrating over radius, using ${\int }_0^{\infty }r{e}^{-r^2/2}dr=1$, gives $\boxed{f_{\Theta }(\theta )=1/(2\pi )}$ on the angle interval. The joint density factors as the product of these marginals, so $\boxed{R\text{ and }\Theta \text{ are independent}}$. The arbitrary angle assigned at $R=0$ is irrelevant because that event has probability zero.