Finding the joint density of $Z=X+Y$ where $X\in U(0,1), Y\in U(0,\alpha)$ I'm trying to find the joint density of $Z=X+Y$ where $X\in U(0,1), Y\in U(0,\alpha)$
Here $U$ is the uniform distribution.
The method I use i to introduce an auxilary variable $W=X$ and then use the transformation theorem according to which we have $f_{ZV}(z,v) = f_{XY}(v,z-v)|J| = \alpha^{-1}$
Now if I just could integrate with respect to $w$ I would be done, but I don't know which regions to integrate over. At least $v$ shold range from $0$ to $1$, and $z$ from $0$ to $1+\alpha$, but which is the third? How do I picture this? Also I think I may need to take into account whether $\alpha$ is greater or less than $1$? 
Thanks in advance!
 A: Suppose $\alpha< 1$ (for $\alpha>1$, it is similar.)
To calculate the probability (density) of a number $t$ with $0<t<1+\alpha$, we have to evaluate
$$
\int_{x=0}^{t}P_1(x)P_2(t-x)\,dx
$$
where $P_1$ is the uniform distribution $(0,1)$ and $P_2$ is the uniform distribution on $(0,\alpha)$. When $0< t a$, we get
$$
\int_{x=0}^{t}P_1(x)P_2(t-x)\,dx=\int_{x=0}^{t}1\cdot \frac 1\alpha\,dx=\frac t\alpha
$$
When $\alpha<t<1 $, we get
\begin{align}
\int_{x=0}^{t}P_1(x)P_2(t-x)\,dx&=
\int_{x=0}^{t-\alpha}P_1(x)P_2(t-x)\,dx+\int_{x=t-\alpha}^{t}P_1(x)P_2(t-x)\,dx\\
&=
\int_{x=0}^{t-\alpha}1\cdot 0\,dx+\int_{x=t-\alpha}^{t}1\cdot \frac 1\alpha\,dx\\
&=0+\alpha\frac 1\alpha=1
\end{align}
When $1<t<1+\alpha$, we get
\begin{align}
&\int_{x=0}^{t}P_1(x)P_2(t-x)\,dx=\int_{x=0}^{t-\alpha}P_1(x)P_2(t-x)\,dx\\
&+\int_{x=t-\alpha}^{1}P_1(x)P_2(t-x)\,dx+\int_{x=1}^{t}P_1(x)P_2(t-x)\,dx\\
&=\int_{x=0}^{t-\alpha}1\cdot0\,dx+\int_{x=t-\alpha}^{1}1\cdot\frac 1\alpha \,dx+\int_{x=1}^{t}0\cdot\frac1\alpha\,dx\\
&=0+(1-(t-\alpha))\frac 1\alpha+0=\frac{1-t+\alpha}{\alpha}
\end{align}
We have to integrate over three regions here, because there are several critical points: $P_1(x)$ is discontinuous at $x=0$ and $x=1$ and $P_2(x)$ at $0$ and $\alpha$. Because we have $P_x(t-x)$, they are at $x=t$ and $x=t-\alpha$. Now, we sort them:
$$
0<t-\alpha<1<t
$$
Because of the discontinuities, we have to integrate each of the three parts of the interval $[0,t]$ on its own.
We can plot the result:

