= Solution
{c}
For $\alpha=3$, the <divergence> vanishes at every point with $r>0$, so its ordinary <volume integral>, ignoring the undefined single point, is zero. Nevertheless, the outward <flux integral> through the unit <sphere> is still $4\pi$:
$$
\boxed{\iiint_{0<r\leq1}\nabla\cdot\mathbf u\,dV=0,
\qquad\iint_{r=1}\mathbf u\cdot\mathbf n\,dS=4\pi}.
$$
\b[The hypotheses of the divergence theorem fail at the origin.] Indeed $|\mathbf u|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\varepsilon<r<1$, the <divergence theorem> remains valid: the outer <flux> is $4\pi$ and the inner <flux> is $-4\pi$. Unlike <iii>, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent <flux> at the singular point, rather than a failure of the <divergence theorem> under its hypotheses.
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