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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 clearenter image description here

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    $\begingroup$ As I've typically seen it, $\frac{\partial \phi}{\partial n}$ is a shorthand notation for the change in $\phi$ as it moves normal to the surface $S$. They then vectorize this to make it into a gradient. $\endgroup$ – infinitylord Dec 24 '17 at 4:01
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$\int_{S}\frac{\partial\phi}{\partial n}dS=\int_{S} {\nabla \phi \cdot \boldsymbol{n}} dS=\int_{V} {\nabla \cdot\nabla \phi } d\tau =\int_{V} {\nabla^2 \phi } d\tau=\int_{V} {0} d\tau=0$ by the divergence theorem since $\phi$ is harmonic.

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  • $\begingroup$ $\int_{S}\frac{\partial\phi}{\partial n}dS=\int_{S}{\nabla\phi\cdot\boldsymbol{n}}dS$ Please explain this $\endgroup$ – Mohan Sharma Dec 24 '17 at 3:57
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    $\begingroup$ This did not address the questions that the OP asked. $\endgroup$ – infinitylord Dec 24 '17 at 3:58
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    $\begingroup$ $\frac{\partial\phi}{\partial n}$ is the notation defined by $\frac{\partial\phi}{\partial n}=\nabla\phi \cdot n$ $\endgroup$ – Chris kim Dec 24 '17 at 3:59
  • $\begingroup$ What is $\partial n$? is $\partial n$ anywhere related to normal unit vector ? $\endgroup$ – Mohan Sharma Dec 24 '17 at 3:59
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    $\begingroup$ Yes, it means the directional derivative of $\phi$ along the normal vector $n$ of the surface $S$. $\endgroup$ – Chris kim Dec 24 '17 at 4:03
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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 $\text{div}(W)=\dfrac{\partial^{2}\phi}{\partial x_{1}^{2}}+\cdots+\dfrac{\partial^{2}\phi}{\partial x_{n}^{2}}=0$.

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  • $\begingroup$ $\dfrac{\partial\phi}{\partial n}$ means the directional derivative of $\phi$ at $x$ with normal $n_{x}$. $\endgroup$ – user284331 Dec 24 '17 at 4:02

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