Prove that $\int_S n\times r dS=0$ 
If $r$ be the position vector of a point on a closed surface $S$ and $n$ be the unit normal (outward) vector to $S$, then prove that $$\int_S n\times r\,dS=0$$

Attempt:
$r=xi+yj+zk$, $n=\frac{\nabla \phi}{|\nabla \phi|}$, where $\phi$ is the given surface.
Then how to proceed? The form of $\phi $ is not given.
 A: Dotting with an arbitrary vector $v$, we have
\begin{align*}
\int_S \langle n\times r, v\rangle dS&=\int_S\langle r\times v, n\rangle dS\\
&=\int_B \nabla\cdot(r\times v) dV\\
&=\int_B v\cdot(\nabla\times (x, y, z))dV\\
&=\int_B v\cdot 0dV\\
&=0
\end{align*}
where $B$ is the solid enclosed by $S$. So the original integral, which is a vector in $\mathbb{R}^3$, is zero.
A: Useful vector identity: if $V$ is a volume with closed bounding surface $S$,
$$ \int_V \nabla \times F \, dV = \int_S n \times F \, dS. $$
This can be proved in the same way as the divergence theorem, or, as Alex Fok suggests, dotting with the constant vector $v$,
$$ v \dot \int_V \nabla \times F \, dV =  \int_V v \cdot (\nabla \times F) \, dV \\
= \int_V \nabla \cdot ( F \times v ) \, dV \\
 =  \int_S (F \times v) \cdot n \, dS = v \cdot \int_S n \times F \, dS, $$
where the scalar triple product identity
$$ a \cdot (b \times c) = b \cdot (c \times a) $$
and the divergence theorem,
$$ \int_V \nabla \cdot F \, dV = \int_S n \cdot F \, dS $$
have been used.
Applying this to your question, you just have to check that $\nabla \times r=0$.
A: Switch to Einstein Summation Notation and apply the Divergence Theorem,
\begin{align}
\int_{S} \epsilon_{ijk}n_{j}r_{k}dS 
&= \int_{V} \frac{\partial \epsilon_{ijk}r_{k} }{\partial r_{j} } dV\\
&= \int_{V} \epsilon_{ijk} \delta_{kj} dV\\
&= \epsilon_{ijj} \int_{V}dV\\
&= 0
\end{align}
