## Proof of Identity for Intefral of Product of a Function With a Gradient Around a Curve

Theorem
If
$\phi , \: \psi$
are functions and
$S$
is a surface with boundary
$C$
then
$\oint_C \phi \mathbf{\nabla} \psi \cdot d \mathbf{r} = \int \int (\mathbf{\nabla} \phi \times \mathbf{\nabla} \phi ) \cdot d \mathbf{n}$
, with
$C$
$S$
. 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
$\mathbf{F} = \phi \mathbf{\nabla} \psi$
then
$\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times (\phi \mathbf{\nabla} \psi)) \cdot \mathbf{n} dS$

Now use the identity
$(\mathbf{\nabla} \times (\phi \mathbf{\nabla \psi}) = \phi \mathbf{\nabla} \times \mathbf{\nabla \psi} + (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} ) = (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} )$

Since
$\phi \mathbf{\nabla} \times \mathbf{\nabla \psi} =0$

We get
$\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \phi \times \mathbf{\nabla \psi} ) \cdot \mathbf{n} dS$

Substitute the last identity into
$\oint_C (\phi \mathbf{\nabla} \psi) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times (\phi \mathbf{\nabla} \psi)) \cdot \mathbf{n} dS$

to get
$\oint_C \phi \mathbf{\nabla} \psi \cdot d \mathbf{r} = \int \int (\mathbf{\nabla} \phi \times \mathbf{\nabla} \phi ) \cdot d \mathbf{n}$
.