If $\phi$ $is$ harmonic in V then show that$\int\int_{S}\frac{\partial\phi}{\partial n}dS=0$ where S is surface enclosing V

Question

QuestionIf $\phi$ $is$ harmonic in V then show that$\int\int_{S}\frac{\partial\phi}{\partial n}dS=0$ where S is surface enclosing V

MY Problem This is a solved problem in my book but i can't understand it on the following points

1.What is $\frac{\partial\phi}{\partial n}$? What is $\partial n$?

I know $\hat{n}$is unit normal to the surface.

2.In the solution $\frac{\partial\phi}{\partial n}$$\mathbf{n}$= $\nabla\phi$This is used.

There is no formula given in the book related to this concept.I can't understand this formula.How is this possible?

Edit I am going add a picture of the solution,to make it more clear

Answer 1

$\int_{S}\frac{\partial\phi}{\partial n}dS=\int_{S} {\nabla \phi \cdot \boldsymbol{n}} dS=\int_{V} {\nabla \cdot\nabla \phi } d\tau =\int_{V} {\nabla^2 \phi } d\tau=\int_{V} {0} d\tau=0$ by the divergence theorem since $\phi$ is harmonic.

Answer 2

By divergence formula we have \begin{align*} \int_{V}\text{div}(W)dx=\int_{S}W\cdot n_{x}dS(x), \end{align*} put $W=\nabla\phi=\left(\dfrac{\partial\phi}{\partial x_{1}},...,\dfrac{\partial\phi}{\partial x_{n}}\right)$, then $\dfrac{\partial\phi}{\partial n}=\nabla\phi\cdot n_{x}$, and $\text{div}(W)=\dfrac{\partial^{2}\phi}{\partial x_{1}^{2}}+\cdots+\dfrac{\partial^{2}\phi}{\partial x_{n}^{2}}=0$.