Normal Distribution, Probability a point is inside of a circle in $\mathbb{R^2}$

Question

We have $X$~$Normal(0,1)$,$Y$~$Normal(0,3)$. What is the Probability that $M \in [(x,y) \in \mathbb{R^2} : 1 \leq 3x^2 + y^2 \leq 3]$? \n

What I did, I set $\tilde{Y} := \frac{Y}{\sqrt{3}},\tilde{Y}$~$Normal(0,1)$. Now we have $\frac{1}{3} \leq x^2 + \tilde{y}^2 \leq 1$, a donut. I'm lost here, my professor hinted polar co-ordinates but I don't really see how, because if I set

$$ X:= \cos\theta, Y:= \sin\theta, $$ $\theta$ would have 2 different distribution function.

Answer 1

The probability density for $\theta$ is uniform on $[0, 2\pi)$ because $(X,\tilde Y)$ is radially symmetric. So you need to calculate the density of $R = \sqrt{X^2 + \tilde Y^2}$ and compute $\Pr[1 \le R \le 3]$ or equivalently, $\Pr[1 \le R^2 \le 9]$. Hint: $R^2$ is a chi-squared random variable with $2$ degrees of freedom.

Answer 2

You need to integrate over the green region: