## Integrating a Function About a Curve

Theorem
If
$\phi$
is a function and
$S$
is a surface with boundary
$C$
then
$\oint_C \phi d \mathbf{r} = \int \int d \mathbf{S} \times (\mathbf{\nabla} \phi)$
, with
$C$
taken anticlockwise about
$S$
. Proof
Stoke's Theorem states
$\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS$

Let
$\mathbf{F} = \phi \mathbf{a}$
where
$\mathbf{a}$
is a constant vector then
$\oint_C ( \phi \mathbf{a} ) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times ( \phi \mathbf{a})) \cdot \mathbf{n} dS$

We can take the dot product out as a factor from the left hand side
$\oint_C (\mathbf{a} \phi ) \cdot d \mathbf{r} = \mathbf{a} \cdot \oint_C \phi d \mathbf{r}$

On the right hand side, since
$\mathbf{a}$
is a constant vector,
$\mathbf{\nabla} \times (\phi \mathbf{a})= (\mathbf{\nabla} \phi ) \times \mathbf{a}$
so the surface integral becomes
$\int \int_S (\mathbf{\nabla} \times ( \phi \mathbf{a})) \cdot \mathbf{n} dS = \int \int_S (\mathbf{\nabla} \phi) \times ( \mathbf{a})) \cdot \mathbf{n} dS$

Now use the identity
$(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})= \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b} )$
to get
$\int \int_S (\mathbf{\nabla} \phi) \times ( \mathbf{a})) \cdot \mathbf{n} dS = \int \int_S \mathbf{a} \cdot (\mathbf{n} \phi ( \mathbf{\nabla} \phi)) dS = \mathbf{a} \cdot \int \int_S d \mathbf{S} \times (\mathbf{\nabla} \phi)$

Since
$\mathbf{a}$
is arbitrary,
$\oint_C \phi d \mathbf{r} = \int \int d \mathbf{S} \times (\mathbf{\nabla} \phi)$