= Solution
In <cylindrical coordinates>, the surface has radius $r=\sqrt2z$. It is the lateral surface of a <right circular cone>, with vertex at the origin and axis along positive $z$. Its height is $h$ and the limiting upper rim has radius $\sqrt2h$; the horizontal disk at that rim is absent. The half-angle from the vertical satisfies
$$
\boxed{\tan\alpha=\sqrt2,\qquad \alpha\simeq54.7^\circ.}
$$
The upper rim is excluded by $z<h$, which does not affect any convergent <surface integral>. The vertex likewise contributes no area.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1-q1-conical-flux.png]
{title=The open lateral cone and its auxiliary closing cap}
{description=Original sketch in units of $h$. The dashed upper rim bounds an auxiliary disk used only to apply the divergence theorem. The arrow shows the outward lateral normal, which has a downward component.}
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