Everything starts with a good sketch:
The particular object that we have is called a frustum.
As our solid is symmetric with respect to rotation about the $z$-axis, you are correct that $0 \leq \theta \leq 2 \pi$.
We have two choices: bound $z$ in terms of $r$, or bound $r$ in terms of $z$.
If we fix a value for $z \in [0,1]$, then $r$ ranges from $0$ out to the edge of a cone at a value of...
$z + 2 = r$.
If we choose to bound $z$ in terms of $r$, then we will need to divide our integral into two regions---a central central cylinder (where the $z$ values range from $0$ to $1$), and an outer section where the $z$ values start at the surface of the cone and extend upwards to $z=1$.
We thus have two different ways to set up a triple integral for the volume, as well as geometric formula. Make sure that you get the same value for all of them!