Convolution of two independent uniform variables Question: Let's say $X$ and $Y$ are two independent random variables with $Uniform (0,3)$ distribution. What is the probability density function of $X+Y$?
In solution, it is told to use convolution method by many sources since the pdf of the sum of two independent random variables $(X+Y)$ is the convolution of the individual pdfs.
However, I couldn't solve this problem by looking at $Uniform (0,1)$ version solutions of this question. What I've done by now is; 
$fx+y(a) = ∫_0^3 fx(a-y)*fy(y)dy$

$= 1/3 *∫_0^3 fx(a-y)dy$

If what I've found until here is true (I have doubts about upper and lower limits of the integral), how should I determine the limits of pdf of X+Y? Some sources assume $a=.5$ and $a=1.5$ and solve it that way; while other sources assume $0<a<1$ and $1<a<2$ and solve it that way. In both solutions, they don't explain how they find upper and lower limit of the integral. Fully solve the problem given above to clearly explain.

Let $Z = X + Y$, then probability distributionn function of $Z$ is defined as
$$
\begin{aligned}
F_Z(z) &= \text{Pr}(Z \leq z) = \text{Pr}(X + Y \leq z) = \text{Pr}(X \leq z - Y) = \int\limits_{-\infty}^{+\infty}F_X(z-y)f_Y(y)dy = \\
&= \frac{1}{3}\int\limits_{0}^{3}F_X(z-y)dy = 
\left|
\begin{array}{l}
u = z-y \Leftrightarrow du = -dy \\
y = 0 \Leftrightarrow u = z \\
y = 3 \Leftrightarrow u = z-3
\end{array}
\right| = \frac{1}{3}\int\limits_{z-3}^zF_X(u)du = \\ 
& = \left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{1}{3}\int\limits_{0}^z\frac{u}{3}du = \frac{z^2}{18}, & 0 < z \leq 3 \\
\frac{1}{3}\int\limits_{z-3}^3\frac{u}{3}du + \frac{1}{3}\int\limits_{3}^zdu=  
\frac{1}{2}-\frac{(z-3)^2}{18} + \frac{z-3}{3} = -\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right.
\end{aligned}
$$
So, we have that
$$
F_Z(z)=\left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{z^2}{18}, & 0 < z \leq 3 \\
-\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right. \Rightarrow f_Z(z) = F_Z'(z) = 
\left\{\begin{array}{rl}
0, & z \leq 0 \\
\frac{z}{9}, & 0 < z \leq 3 \\
-\frac{z-6}{9}, & 3 < z \leq 6 \\
0, & z > 6
\end{array}
\right.
$$
