I think I'm missing something basic here, so any help is appreciated!
Take $P_X(x) \sim \text{Uniform}(0, 1)$ and $P_Y(y) \sim \text{Normal}(0, \sigma^2)$. The joint probability distribution is
$$P_{XY}(x, y) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-y^2/(2\sigma^2)}$$
for $x \in [0, 1]$. We wish to transform this distribution to polar coordinates, $r = \sqrt{x^2 + y^2}$, $\theta = \text{arctan}(y/x)$. The determinant of the polar transformation is $|r|$, so my intuition is that the transformed distribution is
$$P_{R\Theta}(r, \theta) = \frac{r}{\sqrt{2\pi\sigma^2}}e^{-(r\sin\theta)^2/(2\sigma^2)}$$
for $0 < r < 1/\cos\theta$ and $\theta \in [-\pi/2, \pi/2]$. It is clear, however, that the conditional probability $P_{R | \Theta}(r | \theta)$ does not match the expression in the Cartesian form. Take, for example, $\theta = \pi/2$. We have $x = r\cos\theta = 0$ and $y = r\sin\theta = r$, so
$$P_{XY}(x = 0, y = r) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-r^2/(2\sigma^2)}$$
for $0 < r < 1/cos\theta$, which is the expected half-normal distribution. In the polar form, however, we get (ignoring the normalization)
$$P_{R|\Theta}(r | \theta = \pi/2) \propto \frac{r}{\sqrt{2\pi\sigma^2}}e^{-r^2/(2\sigma^2)}$$
for $0 < r < 1/cos\theta$, which is a Rayleigh distribution.
Where's my mistake? Thanks in advance!
Update
I worked through the normalization term:
$$P_\Theta(\theta) = \int_{0}^{1/\cos\theta}P_{R\Theta}(r,\theta)dr = \frac{\sigma^2}{(\sin\theta)^2\sqrt{2\pi\sigma^2}}\left(1 - e^{-(\tan\theta^2)/(2\sigma^2)}\right)$$
And thus the conditional probability is
$$P_{R|\Theta}(r | \theta) = \frac{P_{R\Theta}(r,\theta)}{P_{\Theta}(\theta)} = \frac{r(\sin\theta)^2}{\sigma^2}\frac{e^{-(r\sin\theta)^2/(2\sigma^2)}}{1 - e^{-(\tan\theta^2)/(2\sigma^2)}}$$
Something still isn't right!
After a night's sleep and some retrospection, I understand why this conditional probability has the form that it does, and it is actually quite simple. This is just another example of Borel's paradox.
$$P_{XY|\Theta}(r\cos\theta, r\sin\theta | \theta = \pi/2) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-r^2/(2\sigma^2)}$$
describes the conditional probability along the $+y$ axis within a thin strip of $x$ and follows the expected half-normal distribution. On the other hand,
$$P_{R|\Theta}(r | \theta=\pi/2) = \frac{r}{\sigma^2}e^{-r^2/(2\sigma^2)}$$
describes the conditional probability along the $+y$ axis within a small wedge of $\theta$, which must grow proportionally to $r$.