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
\[S_2\]
\[\oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r} = \int \int_{S_2} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_2 \]
Adding these gives
\[\oint_{ C} \mathbf{F} \cdot d \mathbf{r} + \oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r} = \int \int_{S_1} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_1 + \int \int_{S_2} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS_2 = \int \int_{S} (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS \]
The left hand side is zero since
\[\oint_{ C} \mathbf{F} \cdot d \mathbf{r} =- \oint_{ \bar C} \mathbf{F} \cdot d \mathbf{r} \]
Hence
\[\int \int \mathbf{\nabla} \times \mathbf{F} dS =0\]