Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the distribution function of $Z=X+Y$ - how to break into cases? 
Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the
  distribution function of $Z=X+Y$ - how to break into cases?

I first found the density functions:
$$
f_x(t) =\begin{cases}
1 && t\in[0,1] \\
0 && else 
\end{cases}
$$
$$
f_y(t) =\begin{cases}
\frac{1}{3} && t\in[-1,2] \\
0 && else 
\end{cases}
$$
Now:
$ F_z(t)=P(Z\leq t)=P(X+Y\leq t)=\int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$
But now I am stuck with breaking the result into cases.
How could it be done? I simply can't understand how to break that into cases when we have two differently distributed variables? 
With one variable I would usually draw the function and then break to cases according to it's behavior, but how could it be done here?
Thanks

Assuming that you know that the sum of two independent random variables is their convolution, we have 
$$F_z(t)= \int_{-\infty}^{\infty}f_X(t-\tau)f_Y(\tau) \ d \tau = \frac{1}{3}  \int_{-1}^{2}f_X(t-\tau)  \ d \tau = \frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx$$
Now, we have cases according to the integral boundaries and the PDF of $X$, namely:
1) If $t+1<0$, i.e. $t<-1$, the integral evaluates to zero.
2)If $t-2>1$, i.e. $t>3$, the integral evaluates to zero.
3) If $t-2<0$ ($t<2$) given that $t+1<1$, i.e. $t<0$ So for $-1 < t < 0$
$$F_z(t)=\frac{1}{3}\int_{0}^{t+1}f_X(x) \ dx = \frac{1}{3}(t+1)$$
4) If $t-2 > 0$ ($t > 2$). So for $2 <t < 3$
$$F_z(t)=\frac{1}{3}\int_{t-2}^{1}f_X(x) \ dx = \frac{1}{3}(3-t)$$
5) Now $0 < t < 2$
$$F_z(t)=\frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx =\frac{1}{3}\int_{0}^{1} 1 \ dx =\frac{1}{3}$$
