The Curl of a Vector in Terms of a Surface Integral

We can write the curl of a vector,
$\mathbf{F}$
, written
$\mathbf{\nabla} \times \mathbf{F}$
in terms of a surface integral:
$\mathbf{\nabla} \times \mathbf{F}= lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_{ \delta S} \mathbf{n} \times \mathbf{F} dS$
, where
$\delta S$
is the surface of the volume
$\delta V$

To show this we use the identity
$\int \int \int_{\delta V} \mathbf{\nabla} \times \mathbf{F} dV = \int \int_{\delta S} \mathbf{n} \times \mathbf{F} dS$

Take the dot product of this identity with
$\mathbf{i} , \: \mathbf{j}, \: \mathbf{k}$
respectively to give
$\int \int \int_{\delta V} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{i} dV = \int \int_{\delta S} (\mathbf{n} \times \mathbf{F}) \cdot \mathbf{i} dS$

$\int \int \int_{\delta V} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{j} dV = \int \int_{\delta S} (\mathbf{n} \times \mathbf{F}) \cdot \mathbf{j} dS$

$\int \int \int_{\delta V} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{k} dV = \int \int_{\delta S} (\mathbf{n} \times \mathbf{F}) \cdot \mathbf{k} dS$

The Mean Value Theorem for Volumes states
$\int \int \int_{\delta V} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{i} dV = \delta V ((\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{i})_{(x_0 , y_0 , z_0 )}$
for some point
$(x_0 , y_0 , z_0 ) \in \delta V$

Le
$\delta V \rightarrow 0$
then
$\mathbf{\nabla} \times \mathbf{F}= lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_{ \delta S} \mathbf{n} \times \mathbf{F} dS$