## Gradient of a Function as a Surface Integral

FIn English, the gradient of a scalar field
$\phi$
,
$\mathbf{\nabla} \phi$
at a point is the rate of change of
$\phi$
in the direction normal to the surface
$S$
defined by
$\phi = CONSTANT$

Mathematically,
$\mathbf{\nabla} \phi = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \cdot \mathbf{n} dS$

To show this use the identity
$\int \int \int_{V} \nabla \phi dV = \int \int_S \phi \mathbf{n} dS$

Taking the dot product with
$\mathbf{i}$
gives
$\int \int \int_{V} \nabla \phi \cdot \mathbf{i} dV = \int \int_S \phi \mathbf{n} \cdot \mathbf{i} dS$

By the Mean Value Theorem for volumes,
$\int \int \int_{\delta V} \mathbf{\nabla} \phi dV =(\mathbf{\nabla} \phi)_{(x_0 , y_0 , z_0)} \delta V \rightarrow (\mathbf{\nabla} \phi)_{(x_0 , y_0 , z_0)} = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \mathbf{n} dS, \: (x_0 , y_0 , z_0) \in \delta V$

Hence
$\mathbf{\nabla} \phi_{(x_0 , y_0 , z_0 )} \cdot \mathbf{i} = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \mathbf{n} \cdot \mathbf{i} dS, \: (x_0 , y_0 , z_0) \in \delta V$

Similarly
$\mathbf{\nabla} \phi_{(x_0 , y_0 , z_0 )} \cdot \mathbf{j} = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \mathbf{n} \cdot \mathbf{j} dS, \: (x_0 , y_0 , z_0) \in \delta V$

$\mathbf{\nabla} \phi_{(x_0 , y_0 , z_0 )} \cdot \mathbf{k} = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \mathbf{n} \cdot \mathbf{k} dS, \: (x_0 , y_0 , z_0) \in \delta V$

Then
$\mathbf{\nabla} \phi = (\mathbf{\nabla} \cdot \mathbf{i} ) \mathbf{i} + (\mathbf{\nabla} \cdot \mathbf{j} ) \mathbf{j} + (\mathbf{\nabla} \cdot \mathbf{k} ) \mathbf{k}$

$\mathbf{n} = (\mathbf{n} \cdot \mathbf{i} ) \mathbf{i} + (\mathbf{n} \cdot \mathbf{j} ) \mathbf{j} + (\mathbf{n} \cdot \mathbf{k} ) \mathbf{k}$

Hence
$\mathbf{\nabla} \phi = lim_{\delta V \rightarrow 0} \frac{1}{\delta V} \int \int_S \phi \cdot \mathbf{n} dS$