This is the sketch i made. Maybe its wrong, because i think it gives a very difficult integral like this. I'll look into it on the meanwhile.
If i'd put that answer on the test, i'd have 0.
The order of the integration must be the one they want.
Anyway i found i drew the plane wrong. Both intersections of the plane with the cilinder should be parallel to the x axis. So that makes it much easier.
Now that i drew it well, i think it's this. Correct me if im wrong please
$\displaystyle \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz$