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Away from the vertex, use the defining function $H=z^2-x^2-y^2$. Its gradient is normal to the smooth part of the double circular cone, andon the surface. Hence the field is tangent at every regular point of the cone. At the origin the cone has no unique tangent plane, but $\mathbf{F}(0)=0$, which belongs to its tangent cone. An equivalent argument, valid also at the vertex, uses the integral curves of a vector field: $\dot{\mathbf{x}}=\mathbf{x}$ has solution $\mathbf{x}(t)=e^t\mathbf{x}(0)$, and these dilations preserve $H=0$. This is an example of radial dilation tangency to homogeneous zero sets.

At every regular point of the double circular cone, its normal is perpendicular to $\mathbf{F}$ by part (a). Therefore the integrand of the flux integral vanishes pointwise:The vertex, if included, is a single point of zero surface area and does not alter the surface integral. The answer is independent of both the subset and its orientation. Tangency, rather than a cancellation between different parts of the surface, explains the zero flux.

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.