## Proof That Integral of Curl of a Vector Field Dot Product Normal is Zero Over a Closed Surface

Theorem
Let

$S$
be any closed surface and let
$\mathbf{F}$
be any vector fields with differentiable components.
Then
$\int \int \mathbf{\nabla} \times \mathbf{F} dS=0$

Proof
Cut the surface into two surfaces
$S_1 , \; S_2$
with the same boundary
$C$
.
Apply Stoke's Theorem
$\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS$
to
$S_1$
and
$S_2$
separately. The boundary of each surface traverses
$C$
in the opposite sense -
$C, \: \bar C$
being anticlockwise and clockwise respectively.
For
$S_1$

$\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_{S_1} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_1$

For
$S_2$

$\oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r} = \int \int_{S_2} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_2$

$\oint_{ C} \mathbf{F} \cdot d \mathbf{r} + \oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r} = \int \int_{S_1} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_1 + \int \int_{S_2} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_2 = \int \int_{S} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS$
$\oint_{ C} \mathbf{F} \cdot d \mathbf{r} =- \oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r}$
$\int \int \mathbf{\nabla} \times \mathbf{F} dS =0$