$X$ and $Y$ has uniform distribution. Find $(X-Y)^2$

Question

Let $X$ and $Y$ be a independent random variable with uniform distribution on the $[-1,1]$. Let $Z=(X-Y)^2$. Find :

1. Expected value: $ E(Z)=E(X^2)-2E(X)E(Y)+E(Y^2)=\frac{4}{3}$

2. Probablity density function:

Here, I have problem. I think, that I can write :

$P((X-Y)^2 \le t)$=\begin{cases} 0 &\text{for } t \le 0\\P((X-Y)\le \sqrt{t}) &\text{for } 0<t\le1 \\1&\text{for}\ t\ge1 \end{cases} I'm not sure what to do next..