I have to give the distribution of $Z_1^2 + Z_2^2$ knowing that $Z_1$ and $Z_2$ are independent standard normal variables.
I've found that $Z_1^2 + Z_2^2$ it is typically Chi-squared distributed and hence it's easy to find it's cdf on the internet, but I fear that I'm missing the core of the answer. Would it be sufficient to say : "Well it is Chi-squared distributed, if you look in your book the cdf of that distribution is this one. The end" ?
Calculate $P(Z_1^2+Z_2^2\le z)$, using the normal density function, then switch to the polar coordinates ( you will integrate on the disc of radius$\sqrt{z}$, and center(0,0)), finally you will get $P(Z_1^2+Z_2^2\le z)=e^{-0.5z}$, which is the cdf of a chi-squared distribution with d=2
More details : F is the cdf of $Z_1^2+Z_2^2$ $F(z)=P(Z_1^2+Z_2^2\le z)=\int_{\mathbb{R^2}}{1_{\{z_1^2+z_2^2\leq z\}}\frac{1}{2\pi}e^{-\frac{1}{2}{(z_1^2+z_2^2)}}dz_1dz_2}$
By changing the coordinates
$=\frac{1}{2\pi}\int_{0}^{2\pi}{\int_{0}^{\sqrt{z}}{e^{-\frac{1}{2}{r^2}}rdrd\theta}}$
Integrating $\theta$ first and finally r, we have the result wanted