Calculate distributions from random vector 
Let $(X,Y)$ be a random vector with $X,Y$ Normal Standard, independent
  and $f_{x,y}(x,y) = \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$. Calculate
  the distribution of $X^2+Y^2$ and $arctan(\frac{Y}{X})$

I don't know how to approach this exercise. I've calculated the marginals of $X$ and $Y$ but i don't know where to go from here. I would appreciate any help. 
 A: The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = \iint_D \frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy.$$
Doing the usual polar coordinates ($x = r \cos \theta$ and $y = r \sin \theta$) change of variable gives:
$$dx\,dy=r\,dr\,d\theta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < \sqrt{z}$ and no restriction on $\theta$) 
$$F_Z(z) = \int_0^{2\pi}\frac{1}{2\pi}\left(\int_0^{\sqrt{z}}re^{-r^2/2}\,dr           \right)\,d\theta = \frac{1}{2}e^{-\frac{z}{2}}$$
The above is the CDF of an exponential distribution with parameter $\lambda = \frac{1}{2}$.
The Phase
Let $\alpha= \arctan \frac{Y}{X}$
Let's compute $$F_{\alpha}(\alpha)=\iint_D \frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx\,dy=r\,dr\,d\theta$$
which yields (keeping in mind that $\arctan \frac{y}{x} < \alpha$ will give $\theta < \alpha$ and no restriction on $r$ this time)
$$F_{\alpha}(\alpha) = \int_0^{\alpha}\frac{1}{2\pi}\left(\int_0^{\infty}re^{-r^2/2}\,dr           \right)\,d\theta = \frac{1}{2}e^{-\frac{z}{2}} = \frac{\alpha}{2 \pi}$$
which is a CDF of a uniform distribution over $[0,2\pi]$.
