As you noted, you are looking for a triple integration, but I am posting another parallel approach. If we consider the whole volume, we have $8$ times of the following solid thing(right on the top). This means that, I consider the first $1/8$ of $\mathbb R^3$ for working. And, I am assuming you can find out why the thin slab is really an square (Just examine the equations of both Cylinder). Moreover, we can move that piece from $x=0..x=3$ and so:
$$V_{1/8}=\int_0^3S_{\small{\text{slab}}}(x)dx=\int_0^3(9-x^2)dx$$
I think this way is easier than doing triple integrals, however, you may not like it. So just keep it for the later and let's wait for your desire solution. :-)
