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

Now $v=z^2$ and $\nabla v=2z{\mathbf{e}}_z$. Since $\nabla u=\hat{\mathbf{r}}/a$, their inner product is $2r{\cos }^2\theta /a$. ConsequentlyAlthough $u=1$ on the boundary, the Laplacian is ${\nabla }^{2}v=2$, 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 ${\epsilon }^4/a$ 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 ${\int }_{V}2r/adV=2\pi a^3$, whose difference is the displayed answer.