Prove that $\int_S n\times r dS=0$

Question

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.

Answer 1

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 $\nabla \times r=0$.

Answer 2

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.