= Solution
Now $v=z^2$ and $\nabla v=2z\mathbf e_z$. Since $\nabla u=\widehat{\mathbf r}/a$, their <inner product> is $2r\cos^2\theta/a$. Consequently
$$
\int_V\nabla u\cdot\nabla v\,dV=\frac2a\int_0^a r^3\,dr\int_0^\pi\cos^2\theta\sin\theta\,d\theta\int_0^{2\pi}d\varphi
=\frac2a\frac{a^4}{4}\frac23(2\pi)=\boxed{\frac{2\pi a^3}{3}}.
$$
Although $u=1$ on the boundary, the <Laplacian> is $\nabla^2v=2$, so $v$ is not a <harmonic function>. There is no contradiction with part (a). The origin singularity in the derivative of $u$ is harmless for this integral: its inner boundary contribution is of order $\varepsilon^4/a$ and tends to zero. More explicitly, <Green's first identity> on the punctured ball gives outer flux $8\pi a^3/3$ and volume term $\int_V2r/a\,dV=2\pi a^3$, whose difference is the displayed answer.
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