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 \[C\]
taken anticlockwise about \[S\]
.
ProofStoke'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
\[\oint_C \phi \mathbf{\nabla} \psi \cdot d \mathbf{r} = \int \int (\mathbf{\nabla} \phi \times \mathbf{\nabla} \phi ) \cdot d \mathbf{n} \]
.