Dedicated to great Mark Hennings

Thank you for the dedication! If $\phi = -3xz^2\mathbf{i} + z^3\mathbf{k}$ , and if $T$ is the closed surface obtained by adding the circular cap $x^2 + y^2 \le 4$ , $z=4$ to $S$ , then $\nabla\cdot \phi \,=\, -3z^2 + 3z^2 = 0$ , so the Divergence Theorem gives $\oint_T \phi \cdot d\mathbf{S} \; = \; 0$ The outward-pointing unit normal on the circular cap is $\mathbf{k}$ , and the outward-pointing area element on the surface $S$ is what the question writes as $-d\mathbf{A}$ . Thus $0 \; = \; \int\int_{x^2+y^2 \le 4} 4^3\,dx\,dy - \iint_S \phi \cdot d\mathbf{A} \; = \; 256\pi - \iint_S \phi \cdot d\mathbf{A}$ making the answer $\sqrt{256} = \boxed{16}$ .