Probability density of transformation of 2D normally distributed vector

Question

let (X,Y) be a 2 dimensional normally distributed vector, with X,Y~N[0,1] independently.

Let U=X$^2$+Y$^2$. find the density function f$_u$(u) for u$>$0.

Answer by book: U~exp(0.5)

I tried creating a second transformation V(X,Y) and finding f$_u,_v$ using the density transformation formula,and than getting rid of v using integral. All the different transformations I tried led me to the same integral, that seems to lead me closer to the answer but I can't get rid of v (integral d0esn't converge)

Thanks in advance

Answer 1

Hint: Find the CDF of $U$ by computing $$P(U \leq u) = P(X^2+Y^2 \leq u) = \iint _{\{(x,y) : x^2+y^2 \leq u\}} f_{X,Y}(x,y) \ dy \ dx$$ and convert this to an integral in polar coordinates. Everything will work out to be the CDF of the exponential distribution with $\lambda = 1/2$.