**Hint:** Split the calculation into two cases: (i) $0\le z\le 1$ and (ii) $1\lt z\le 2$.

**Added:** (i) if $0\le z\le 1$, then $f_X(z-y)=1$ if $0\le y\le z$, and $f_X(z-y)=0$ if $y\gt z$. It follows that
$$\int_0^1 f_X(z-y)\,dy=\int_0^z 1\cdot dy=z$$.

(ii) If $1\lt z\le 2$, then $f_X(z-y)=1$ if $z-1\le y \le 1$, and $f_X(z-y)=0$ elsewhere. It follows that
$$\int_0^1 f_X(z-y)\,dy=\int_{z-1}^1 1\cdot dy=2-z.$$
Thus $f_Z(z)=z$ if $0\le z\le 1$, and $f_Z(z)=2-z$ if $1\le z\le 2$. And for completeness, $f_Z(z)=0$ if $z$ is outside the interval $[0,2]$.

**Remark:** I suspect that the convolution way is in this case effectively no faster than the "slow" way of finding first the cumulative distribution function $F_Z(z)$, and differentiating.

Think of the joint pdf being a cheese-cube flushed against axes in the first octant. Now, imagine the line $y = z- x$ being a long blade cutting the cheese-cube from the top. The part of cheese-cube cut off by the blade on the bottom-left corner is the CDF of $X+Y$. For $z < 1$, the part of cheese-cube cut off by the blade on the bottom-left corner is a triangular prism. For $z > 1$, the shape is different (in this case, it is the cube with a triangular prism taken out). Hopefully, this helps you see why the integral needs to be split at $z=1$.

## Best Answer

You want to find the volume of the region described by the following inequalities:

$$0 \leq x, y, z \leq 1$$ $$x/y < z$$

One way to solve this is to note that $x / y < z$ is equivalent to $x < yz$. Note that since $y, z \leq 1$, we have $yz \leq 1$.

$$0 \leq x < yz$$ $$0 \leq y, z \leq 1$$

So the integral in question will be $\int_0^1 \int_0^1 \int_0^{yz} dx dy dz = \frac{1}{4}$.