= 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> $r$. Consequently
$$
\boxed{f_{R,\Theta}(r,\theta)=\frac r{2\pi}e^{-r^2/2},\quad r\geq0,\ 0\leq\theta<2\pi,}
$$
with <density> zero outside this range. Integrating over the angle gives $\boxed{f_R(r)=re^{-r^2/2}}$, the <Rayleigh distribution> density. Integrating over radius, using $\int_0^\infty re^{-r^2/2}dr=1$, 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.
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