Rayleigh distribution with unequal variances Suppose we have two independent, uncorrelated random variables $X\sim N\left(0,a^2\right)$ and $Y\sim N\left(0,b^2\right)$ (i.e. $X$ and $Y$ are Normally distributed with mean 0 and standard deviations $a$ and $b$ respectively.)
How do I find the probability that $\sqrt{X^{2}+Y^{2}}\le r$, where r is a positive real number?
I know that if the two normals had the same standard deviations, then I would use the Rayleigh CDF to answer my question. But when the standard deviations are unequal, I can't seem to find much on how to obtain this probability.

In this case the random variable $X^2+Y^2$ is a weighted sum of two chi-squared random variables each with one degree-of-freedom.  There is no closed form distribution for the probability of interest in the general case where $a \neq b$.  To facilitate our analysis, let $S \equiv Y^2/b^2 \sim \text{ChiSq}(1)$.  We can then write the probability of interest as:
$$\begin{align}
\mathbb{P}(\sqrt{X^2+Y^2} \leqslant r)
&= \mathbb{P}(X^2+Y^2 \leqslant r^2) \\[16pt]
&= \mathbb{P}(X^2+ b^2 S \leqslant r^2) \\[8pt]
&= \int \limits_0^\infty \mathbb{P}(X^2+b^2 S \leqslant r^2|S=s) \cdot \text{ChiSq}(s|1) \ ds \\[6pt]
&= \int \limits_0^\infty \mathbb{P}(X^2+b^2 s \leqslant r^2) \cdot \text{ChiSq}(s|1) \ ds \\[6pt]
&= \int \limits_0^\infty \mathbb{P} \bigg( \Big( \frac{X}{a} \Big)^2 \leqslant \frac{r^2-b^2 s}{a^2} \bigg) \cdot \text{ChiSq}(s|1) \ ds \\[6pt]
&= \frac{1}{\sqrt{2} \pi} \int \limits_0^{r^2/b^2} \gamma \Big( \frac{1}{2}, \frac{r^2-b^2 s}{2 a^2} \Big) \cdot \frac{1}{\sqrt{s}} \cdot \exp \Big( -\frac{s}{2} \Big) \ ds, \\[6pt]
\end{align}$$
where $\gamma$ is the lower incomplete gamma function.  You can evaluate this integral numerically to obtain the probability of interest, but it has no closed form in general (so far as I'm aware).
