= Solution
Away from the origin the <gradients> are $\nabla u=\widehat{\mathbf r}/a$ and $\nabla v=-\widehat{\mathbf r}/r^2$. The improper <volume integral> is therefore
$$
\lim_{\varepsilon\downarrow0}\int_{\varepsilon<r<a}-\frac{1}{ar^2}\,dV
=-\frac{4\pi}{a}\lim_{\varepsilon\downarrow0}\int_\varepsilon^a dr=\boxed{-4\pi}.
$$
The function $v=1/r$ is a <harmonic function> only off the origin. It is singular inside the integration volume, so it does not meet part (a)'s regularity hypothesis. On the punctured ball, <Green's first identity> has outer flux $-4\pi$ and inner flux $4\pi\varepsilon/a$: the inner outward <unit normal> is $-\widehat{\mathbf r}$, so $u\partial_n v=(\varepsilon/a)/\varepsilon^2$. Their sum is $-4\pi+4\pi\varepsilon/a$, exactly the integral before taking the limit. Equivalently the distributional <Laplacian> of $1/r$ is $-4\pi\delta_0$, rather than zero throughout the ball. This is <boundary flux from a singular harmonic potential>; it explains why applying part (a) across the origin would be invalid.
Back to article page
