Stokes theorem says:
$\int_C \vec{F}$•$\vec dr$ =$\iint_\sum (\nabla \mathbf x \vec{F})•\hat n$ $d\sigma$
This means
$\int_C \vec{F}$•$\vec dr$ = $\iint_R (\nabla \mathbf x \vec{F})•\hat n$ $||\partial \vec r/du$ $\mathbf x $ $\partial \vec r/dv||$ $ dudv$.
Right?