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 \[S\]
.
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
\[\mathbf{F} = \phi \mathbf{a}\]
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, \[\oint_C \phi d \mathbf{r} = \int \int d \mathbf{S} \times (\mathbf{\nabla} \phi)\]