 Solution

Use the outward normal vector. The vector field is $\mathbf{F}=(xz,2y,-z^2)$, with divergence $\nabla \cdot \mathbf{F}=z+2-2z=2-z$. By the divergence theorem, its outward flux equals the integral of this divergence over the solid cylinder. Each horizontal slice has area $8\pi$, soFor an independent boundary check, the cylindrical side contributes $84\pi$, the top contributes $(-3^2)(8\pi )=-72\pi$, and the bottom contributes zero. Their sum is again $12\pi$. Reversing all normals changes its sign.