how to show: Is $u$ harmonic, so are following identities true Let $V \subset \mathbb{R}^n , 2 \leq n $
be a set, where you can apply Gauß's Theorem.
To show: Is $ u \in C^{(2)}( \bar{V} ) $ harmonic on $V$, then:
$$ \int_{ \partial V} \frac{ \partial u }{ \partial \nu } do =0 $$
and 
$$ \int_V | \nabla u(x)|^2 dx = \int_{ \partial V} u \frac{ \partial u }{\partial \nu} do $$
where $ \nu $ is the unit normal on $ \partial V $
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Probably it's a bit remodelling, but I still can not get around.
Okay, so the Gauß's Theorem says
$$ \int_{ \Omega} \nabla v dx = \int_{ \partial \Omega} v n ds $$
and If $u \in C^{(2)} $  is harmonic then :
$$ \Delta u = \frac{ \partial^2 u}{ \partial x^2} + \frac{ \partial^2 u}{ \partial y^2}=0 $$
 A: Recall that
$\Delta u = \nabla^2 u = \nabla \cdot \nabla u; \tag 1$
if $u$ is harmonic on $\bar V$, then
$\Delta u = \nabla^2 u(x) = 0, \; x \in \bar V; \tag 2$
we may apply the divergence theorem of Gauss to the integral
$\displaystyle \int_{\partial V} \dfrac{\partial u}{\partial \nu} \; do; \tag 3$
on $\partial V$ we have
$\dfrac{\partial u}{\partial \nu} = \nabla u \cdot \mathbf n, \tag 4$
where $\mathbf n$ is the outward-pointing unit normal vector field on $\partial V$; then using the divergence theorem we find
$\displaystyle \int_{\partial V} \dfrac{\partial u}{\partial \nu} \; do = \displaystyle \int_{\partial V} \nabla u \cdot \mathbf n \; do  = \int_V \nabla \cdot \nabla u \; dV = \int_V \nabla^2 u \; dV = 0, \tag 5$
since $u$ is harmonic on $V$, that is, from (2).
As for the second identity, we use the well-known formula
$\nabla \cdot (u \nabla u) = \nabla u \cdot \nabla u + u \nabla \cdot \nabla u = \vert \nabla u \vert^2 + u\nabla^2u = \vert \nabla u \vert^2 \tag 6$
for harmonic $u$; if we now integrate over $V$ we find
$\displaystyle \int_V \vert \nabla u \vert^2 \; dV = \int_V \nabla \cdot (u \nabla u) \; dV; \tag 7$
another application of the divergence theorem similar to the above yields
$\displaystyle \int_V \nabla \cdot (u \nabla u) \; dV = \int_{\partial V} u \nabla u \cdot \mathbf n \; do = \int_{\partial V} u\dfrac{\partial u}{\partial \nu} \; do; \tag 8$
combining (7) and (8) yields 
$\displaystyle \int_V \vert \nabla u \vert^2 \; dV = \displaystyle \int u \dfrac{\partial u}{\partial \nu} \; do, \tag 9$
the requisite result. $OE\Delta$.
