= Solution
Here $v=z$, so $\nabla v=\mathbf e_z$, while $\nabla u=\widehat{\mathbf r}$ away from the origin. Their <inner product> is $\cos\theta$. The <spherical coordinates> volume element gives
$$
\int_V\nabla u\cdot\nabla v\,dV=\int_0^a r^2\,dr\int_0^\pi\cos\theta\sin\theta\,d\theta\int_0^{2\pi}d\varphi=\boxed{0}.
$$
This 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 $\varepsilon$ resolves this: the inner boundary term $u\partial_n v=-\varepsilon\cos\theta$ has zero angular integral, so it contributes nothing in the limit. Thus the improper <volume integral> has the same zero value.
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