## Proof That Surface Integral of Cross Product Radial Vector With Normal is Zero

Theorem
If
$S$
is a closed surface for a region
$V$
then
$\int \int_S \mathbf{r} \times \mathbf{n} dS =0$
where
$\mathbf{n}$
is the outward normal.
Proof
We use the Identity for volume Integral of the curl of a vector field
$\int \int \int_V ( \mathbf{\nabla} \times \mathbf{F}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{F}) \: dS$

With use t
$\mathbf{F} = \mathbf{f}$
we have
$\int \int \int_V ( \mathbf{\nabla} \times \mathbf{r}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{r}) \: dS$

\begin{aligned} \mathbf{\nabla} \times \mathbf{r} &= (\frac{\partial}{\partial x} \mathbf{i}+\frac{\partial}{\partial y} \mathbf{j}+ \frac{\partial}{\partial k} \mathbf{k}) \times (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \\ &= (\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}) \mathbf{i} + (\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}) \mathbf{j} + (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) \mathbf{k} \\ &= \mathbf{0}\end{aligned}

Hence
$\int \int_S \mathbf{r} \times \mathbf{n} dS =0$