Let $Y$ be a continuous random variable with probability density function$$ f_Y(x)=\begin{cases} xe^{-\frac{x^2}2};&x>0\\ 0;&x\le0\end{cases}.$$ Compute $\mathbb{E}(Y)$ and determine the distribution of $Z$, where $Z=Y^2$.
This is what I think is correct for $\mathbb{E}(Y)$:
$$\mathbb{E}(Y)=\int_0^\infty\int_y^\infty xe^{-\frac{x^{2}}{2}} \,\mathrm{d}x \mathrm{d}y=\infty$$
This is not asked in the question but I am guessing the variance of $Y$ will also be infinity? It cannot be finite right?
And for the distribution function of $Z$, I'm a little bit stuck but this is what I have been able to do:
$$F_{Z}(z)=\mathbb{P}(Z \le z)=\mathbb{P}(X \le \sqrt z).$$
Edit: I realized I named my random variable $X$ rather than $Y$. Sorry!
Note that $$E(Y)=\int_{-\infty}^{\infty} xf_Y(x)dx=\int_{0}^{\infty} x^2 e^{-\frac{x^2}{2}}dx$$
For the second part, you are on the right track.
If the support of $Y$ is on $(0,\infty)$ then so is $Z=Y^2$.
We have
$$\begin{align*} F_Z(z) &=P(Z\leq z)\\\\ &=P(Y^2 \leq z)\\\\ &=P(Y \leq \sqrt{z})\\\\ &=F_Y(\sqrt{z})\\ &\vdots \end{align*}$$
where $P(Y^2 \leq z)=P(Y \leq \sqrt{z})$ because $Y$ does not take on negative values.
You need to integrate the PDF of $Y$ to get its CDF