= Solution
{c}
On the unit <sphere>, the outward <unit normal vector> is $\mathbf n=\mathbf r$ and $\mathbf u=\mathbf r$. Hence the outward <flux integral> is
$$
\iint_{r=1}\mathbf u\cdot\mathbf n\,dS
=\iint_{r=1}1\,dS=4\pi.
$$
Using <spherical polar coordinates> and the <divergence> from <ii>, the <volume integral> over the unit <ball> is
$$
\begin{aligned}
\iiint_{r\leq1}\nabla\cdot\mathbf u\,dV
&=4\pi(3-\alpha)\int_0^1r^{2-\alpha}\,dr\\
&=4\pi(3-\alpha)\left[\frac{r^{3-\alpha}}{3-\alpha}\right]_0^1
=\boxed{4\pi}\qquad(\alpha<3).
\end{aligned}
$$
This <improper integral> converges precisely because $2-\alpha>-1$. When the <vector field> is singular or fails to be continuously differentiable at the origin, justify the equality by applying the <divergence theorem> on $\varepsilon<r<1$. The outward <flux> through its inner <sphere> is $-4\pi\varepsilon^{3-\alpha}$, so
$$
\iiint_{\varepsilon<r<1}\nabla\cdot\mathbf u\,dV
=4\pi\bigl(1-\varepsilon^{3-\alpha}\bigr)\longrightarrow4\pi.
$$
Thus \b[the two integrals agree for every $\alpha<3$], with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.
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