Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 11C / b / i / Solution

Here $v=z$, so $\nabla v={\mathbf{e}}_z$, while $\nabla u=\hat{\mathbf{r}}$ away from the origin. Their inner product is $\cos \theta$. The spherical coordinates volume element givesThis agrees with the boundary-constant gradient pairing with a boundary-constant function and a harmonic function: $u$ has constant boundary value $a$ and $v=z$ is a regular harmonic function. For comparison with the boundary value one in part (a), use $u/a$ and multiply back by $a$. Strictly, $u=r$ is not differentiable at the origin. Excising a ball of radius $\epsilon$ resolves this: the inner boundary term $u{\partial }_{n}v=-\epsilon \cos \theta$ has zero angular integral, so it contributes nothing in the limit. Thus the improper volume integral has the same zero value.