Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics V / 2 / a / iii / Solution

Take $v=|\mathbf{x}{|}^2$ in Green's second identity. Since ${\nabla }^{2}u=0$, ${\nabla }^{2}v=6$, $v=1$ on the unit sphere, and ${\partial }_{n}v=2$ there,The final term vanishes by the divergence theorem, because it equals ${\iiint }_B{\nabla }^{2}udV=0$. ThereforeWe assume sufficient boundary regularity for the stated boundary values and integrals, or use the corresponding trace/limit formulation. This is the equality of volume and surface averages for a harmonic function, since the unit ball and sphere have volume $4\pi /3$ and area $4\pi$.