Given $X$ and $Y$ are independent N(0,1) random variables and $Z = \sqrt{X^2+Y^2}$ from the marginal pdf of $Z$ Let $X$ and $Y$ be independent $N(0; 1)$ random variables. Let $Z = \sqrt{X^2+Y^2}$.
(a) Derive the marginal pdf of $Z$ and then using the marginal pdf to compute ${\rm E}[Z^2]$
(b) Can you propose a different way other than that in (a) to compute ${\rm E}[Z^2]$
(c) Compute ${\rm E}[Z]$.
This is the whole question that I was asked. I can do (c) but I don't know how to find the marginal pdf of $Z$ with the given information and can't seem to find any formulas. Any help would be appreciated. 
 A: Using heropup's answer, we have
$$
f_S(s) = \frac{1}{2} e^{-\frac{1}{2}s}, \quad s > 0
$$
and
$$
F_Z(z)  = \Pr[S \le z^2].
$$
Now, we determine the CDF of $S$.
$$
\begin{align}
F_S(s)&=\int_0^sf_S(t)\ dt\\
&=\int_0^s\frac{1}{2} e^{-\frac{1}{2}t}\ dt,\quad\text{let }\ u=-\frac{1}{2}t\quad\Rightarrow\quad dt=-2\ du\\
&=1-e^{-\frac{1}{2}s},
\end{align}
$$
then
$$
F_Z(z)=F_S(z^2)=1-e^{-\frac{1}{2}z^2}
$$
and
$$
f_Z(z)=\frac{d}{dz}F_Z(z)=\frac{d}{dz}\left(1-e^{-\frac{1}{2}z^2}\right)=z\ e^{-\frac{1}{2}z^2}.
$$
Hence
$$
\text{E}\left[Z^2\right]=\int_0^\infty z^2f_Z(z)\ dz=\int_0^\infty z^3e^{-\frac{1}{2}z^2}\ dz\tag1
$$
and
$$
\text{E}\left[Z\right]=\int_0^\infty zf_Z(z)\ dz=\int_0^\infty z^2e^{-\frac{1}{2}z^2}\ dz.\tag2
$$
$(1)$ and $(2)$ are Gaussian integrals, where
$$
\int_0^\infty z^n e^{-az^2}\ dz =
\left\{ \begin{array}{l l}
\frac{(n-1)!!}{2\cdot(2a)^\frac{n}{2}}\sqrt{\frac{\pi}{a}} & \quad \text{for $n$ even}\\
\\
\frac{\left(\frac{1}{2}(n-1)\right)!}{2\cdot(a)^\frac{n+1}{2}}
& \quad \text{for $n$ odd.}
\end{array} \right.\tag3
$$
About double factorial $(!!)$, you may refer to this link. Thus, using $(3)$, we obtain
$$
\text{E}\left[Z^2\right]=\frac{\left(\frac{1}{2}(3-1)\right)!}{2\cdot\left(\frac{1}{2}\right)^\frac{3+1}{2}}=2
$$
and
$$
\text{E}\left[Z\right]=\frac{(2-1)!!}{2\cdot\left(2\cdot\frac{1}{2}\right)^\frac{2}{2}}\sqrt{\frac{\pi}{\left(\frac{1}{2}\right)}} =\sqrt{\frac{\pi}{2}}.
$$
For a different way to compute $\text{E}\left[Z^2\right]$, you may refer to this link. I hope this help.
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
A: If $X_1, X_2, \ldots, X_n$ are independent and identically distributed standard normal (mean 0, variance 1) random variables, then $$S = X_1^2 + X_2^2 + \cdots + X_n^2$$ is a chi-squared random variable with $n$ degrees of freedom.  The probability density function of $S$ is given by $$f_S(s) = \frac{s^{n/2-1} e^{-s/2}}{\Gamma(n/2)2^{n/2}}, \quad s > 0.$$  In your case, $n = 2$, so $$f_S(s) = \frac{1}{2} e^{-s/2}, \quad s > 0.$$  This also happens to be an exponential distribution with mean $2$.  Now, we are given $Z = \sqrt{S}$.  So to find the density for $Z$, we consider the cumulative distribution function:  $$F_Z(z) = \Pr[Z \le z] = \Pr[\sqrt{S} \le z] = \Pr[S \le z^2].$$  Calculate this last probability and then take the derivative to obtain the density for $Z$.  Then compute an appropriate integration.  Can you do the rest?
A: Hint: You may want to look at Rayleigh Distribution.
