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\[ \int \int \int_V ( \mathbf{\nabla} \times \mathbf{r}) \: dV = \int \int_S ( \mathbf{n} \times \mathbf{r}) \: dS\]
\[\begin{equation} \begin{aligned} \mathbf{\nabla} \times \mathbf{r} &= (\frac{\partial}{\partial x} \mathbf{i}+\frac{\partial}{\partial y} \mathbf{j}+ \frac{\partial}{\partial k} \mathbf{k}) \times (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \\ &= (\frac{\partial z}{\partial y} - \frac{\partial y}{\partial z}) \mathbf{i} + (\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x}) \mathbf{j} + (\frac{\partial y}{\partial x} - \frac{\partial x}{\partial y}) \mathbf{k} \\ &= \mathbf{0}\end{aligned} \end{equation}\]
Hence
\[\int \int_S \mathbf{r} \times \mathbf{n} dS =0\]