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

\(\displaystyle \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz\)