Let
\[\mathbf{F}\]
be a vector field with differentiable components defined on a surface \[S\]
, such that \[\mathbf{\nabla} \times \mathbf{F}\]
is tangential to \[S\]
everywhere on \[S\]
.If
\[C\]
is the border of \[S\]
then \[\oint_C \mathbf{F} \cdot d \mathbf{r} =0\]
Proof
Stoke's Theorem states
\[\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS \]
Since
\[\mathbf{\nabla} \times \mathbf{F}\]
is tangential to \[S\]
everywhere, \[\mathbf{\nabla} \times \mathbf{F}=0\]
everywhere on \[S\]
.Hence
\[\oint_C \mathbf{F} \cdot d \mathbf{r} =0\]
.