Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 3 / 9A / d

Close the truncated two-sheet double circular cone by the lateral circular cylinder $r=1$, $-1\le z\le 1$. Together these form the boundary of the solid in part (c); no planar caps are required because the cone meets the circular cylinder along both end circles. On the circular cylinder the outward unit normal is $(\cos \theta ,\sin \theta ,0)$, so $\mathbf{F}\cdot \mathbf{n}=1$, and $dS=d\theta dz$. Its flux integral isThe divergence theorem now gives ${\int }_S\mathbf{F}\cdot d\mathbf{S}+4\pi =4\pi$, using the volume integral from part (c). Henceagreeing with tangency. The isolated vertex can be handled by excising a ball of radius $\epsilon$ and passing to the limit: since $|\mathbf{F}|=O(\epsilon )$ there and the added area is $O({\epsilon }^2)$, its extra flux is $O({\epsilon }^3)$ and tends to zero.