Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 1

For a continuously differentiable vector field $\mathbf{G}$ on a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, the divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus total outward flux through a closed surface equals the integral of the local divergence inside it. A nonclosed surface first needs a suitable closure, with the added surface's flux accounted for separately.

In cylindrical coordinates, the surface has radius $r=\sqrt{2}z$. 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 $\sqrt{2}h$; the horizontal disk at that rim is absent. The half-angle from the vertical satisfiesThe upper rim is excluded by $z<h$, which does not affect any convergent surface integral. The vertex likewise contributes no area.

Close the cone with the horizontal disk $D$ at height $h$, and let $V$ be the enclosed solid. The divergence isThe term $2y$ integrates to zero by reflection symmetry of each circular cross-section. Using the divergence theorem and cylindrical coordinates, the closed-surface flux isThe outward normal on $D$ is $+\hat{\mathbf{z}}$, so its flux is its area times $G_z=h$:Subtracting this cap contribution gives the required lateral flux:This is conical flux by closing with a cap. The normal convention matters: on the lateral cone the outward normal points radially outwards and partly downwards, so the $G_z$ contribution to the lateral flux is negative. To justify applying the theorem at the vertex, truncate a small tip and take its size to zero; the extra flux vanishes because $\mathbf{G}$ is regular there.