# Surface Integral of the Partial Derivative of a Harmonic Function

Assume that $V$ is a solid in $\mathbb{R}^3$ which is bounded by a surface $S$ whose normal is $\overrightarrow{n}$ and $f:V \rightarrow \mathbb{R}^3$ is a harmonic function on $V$.
Show that $$\iint_S \frac{\partial{f}}{\partial{n}} \,dS = 0$$

My main issue with this question is that I have no idea how to deal with $\frac{\partial{f}}{\partial{n}}$, once I understand what this is referring to I'm pretty sure I'll be able to do it. All help appreciated!

• Is this a UoB MVA question? If so, it is not examinable – anon May 5 '15 at 15:27
• is this the university of Birmingham uk? – Chinny84 May 5 '15 at 15:29
• Yes- University of Birmingham – anon May 5 '15 at 15:31

$\frac{\partial f}{\partial n}$ is shorthand for $(\nabla f) \cdot n$, i.e. the directional derivative in the direction $n$. Really, what this question wants you to do is use the divergence theorem, $$\int_V \nabla \cdot F \, dV = \int_S F \cdot n \, dS,$$ where here you have $F=\nabla f$, and the left-hand side is the integral of the Laplacian.