= Solution
Close the truncated two-sheet <double circular cone> by the lateral <circular cylinder> $r=1$, $-1\le z\le1$. 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> is
$$
\int_{-1}^1\int_0^{2\pi}1\,d\theta\,dz=4\pi.
$$
The <divergence theorem> now gives $\int_S\mathbf F\cdot d\mathbf S+4\pi=4\pi$, using the volume integral from part (c). Hence
$$
\boxed{\int_S\mathbf F\cdot d\mathbf S=0,}
$$
agreeing with tangency. The isolated vertex can be handled by excising a ball of radius $\varepsilon$ and passing to the limit: since $|\mathbf F|=O(\varepsilon)$ there and the added area is $O(\varepsilon^2)$, its extra flux is $O(\varepsilon^3)$ and tends to zero.
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