= 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 $8\pi$, so
$$
\boxed{\oint_{\partial V}\mathbf F\cdot\hat{\mathbf n}\,d\sigma=8\pi\int_0^3(2-z)\,dz=12\pi.}
$$
For 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.
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