Here is another way to proceed that avoids that difficulty. We know that have a joint bivariate normal distribution and that individually have (univariate) normal distributions with mean 0 and standard deviation .
so are uncorrelated, hence independent.
has a chi-square distribution with 2 degrees of freedom, and we can use this fact to compute
But your solution is much, much better and is much more tractable. Thank you for showing it.
My only 'concern' is that a covariance of zero doesn't in general imply independence (although independence does imply a covariance of zero) ..... This is a small detail that can be left for the OP to worry about (s/he should know that there's a thereom about multivariate normal distributions, covariance and independence .....)