## Proof of Identity For Integral of Cross Product of Vector Field With Tangent Around Closed Curve

Theorem
If
$\mathbf{F}$
is a vector field with differentiable components then
$\oint_C d \mathbf{r} \times \mathbf{F}= \int \int_S (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS$
and
$C$
is the boundary of
$S$

Proof
Stoke's Theorem states
$\oint_C \mathbf{A} d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{A}) \cdot \mathbf{n} dS$

Let
$\mathbf{A} = \mathbf{F} \times \mathbf{c}$
(1) where
$\mathbf{c}$
is a constant vector, then,br>
$\oint_C (\mathbf{F} \times \mathbf{c}) \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times ( \mathbf{F} \times \mathbf{c})) \cdot \mathbf{n} dS$

Now use the identity
$\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) =(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$

and the properties of the dot product to rewrite this as
$\oint_C (\mathbf{F} \times \mathbf{c}) \cdot d \mathbf{r} = \oint_C d \mathbf{r} \cdot (\mathbf{F} \times \mathbf{c}) = \oint_C (d \mathbf{r} \times \mathbf{F}) \cdot \mathbf{c} = \oint_C \mathbf{c} \cdot ( d \mathbf{r} \times \mathbf{F}) = \mathbf{c} \cdot \oint_C ( d \mathbf{r} \times \mathbf{F})$
(2)
Npw use the identity
$\mathbf{\nabla} \times (\mathbf{a} \times \mathbf{b})=(\mathbf{b} \cdot \mathbf{\nabla} ) \mathbf{a} - \mathbf{b} ( \mathbf{\nabla} \cdot \mathbf{a}) - (\mathbf{a} \cdot \mathbf{\nabla} ) \mathbf{b} + \mathbf{a} ( \mathbf{\nabla} \cdot \mathbf{b})$

and note that
$\mathbf{\nabla} \cdot \mathbf{c} = 0$
since
$\mathbf{c}$
is a constant vector to get
\begin{aligned} \int \int_S (\mathbf{\nabla} \times (\mathbf{F} \times \mathbf{c})) \cdot \mathbf{n} dS &= \in \int_S ((\mathbf{c} \cdot \mathbf{\nabla}) \mathbf{F} - \mathbf{c} \mathbf{\nabla} \cdot \mathbf{F})) \cdot \mathbf{n} dS \\ &= \int \int)S (\mathbf{c} \cdot \mathbf{\nabla})\mathbf{F} \cdot \mathbf{n} dS - \int \int_S ( \mathbf{c} (\mathbf{\nabla} \cdot \mathbf{F})) \cdot \mathbf{n} dS \\ &= \int \int_S \mathbf{c} \cdot (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S \mathbf{c} \cdot (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \end{aligned}

Substitute this and (2) into (1) to get
\begin{aligned} \mathbf{c} \cdot \oint_C d \mathbf{r} \cdot \mathbf{F} &= \int \int_S \mathbf{c} \cdot (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S \mathbf{c} \cdot (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \\ &= \mathbf{c} \cdot \int \int_S (\mathbf{\nabla} (\mathbf{F} \cdot \mathbf{n}))dS - \int \int_S (\mathbf{n} (\mathbf{\nabla} \cdot \mathbf{F})) dS \\ &= \mathbf{c} \cdot \int \int (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS \end{aligned}

Removing
$\mathbf{c} \cdot {}$
gives
$\oint_C d \mathbf{r} \cdot \mathbf{F} = \int \int (\mathbf{n} \times \mathbf{\nabla} ) \times \mathbf{F} dS$