## Deriving Stoke's Theorem From Green's Theorem

Suppose a curve
$C$
with interior
$B$
defined on
${\mathbb{R}}^2$
has a smooth parametrization
$\mathbf{p}(s) = (p_1 (s) , p_2 (s))$
where
$s$
is the distance along the curve.
The unit tangent to
$C$
is
$\mathbf{T} = ( \frac{p'_1(s)}{\sqrt{p'^2_1(s)+p'^2_2(s)}} , \frac{p'_2(s)}{\sqrt{p'^2_1(s)+p'^2_2(s)}} )$

The unit normal to
$C$
is given by
$\mathbf{n(s)} =\mathbf{T'} = ( \frac{p''_1(s)}{\sqrt{p''^2_1(s)+p''^2_2(s)}} , \frac{p''_2(s)}{\sqrt{p''^2_1(s)+p''^2_2(s)}} )$

The circulation of a force
$\mathbf{F}$
around
$C$
is defined as
$\oint_C F_1 \: dx + F_2 \: dy = \oint_C \mathbf{F} \cdot d \mathbf{r} ds$

$curl \mathbf{F} = (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}) \mathbf{k}$

Hence Green's Theorem
$\oint_C F_1 \: dx + F_2 \: dy = \int_B (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})dx \: dy$

can be written
$\oint_C \mathbf{F} \cdot \mathbf{T} ds = \int \int_B curl \mathbf{F} \; dx \: dy$

This is called Stokes Theorem.