## Stoke's Theorem and 1 - Forms

We can derive Stoke's formula from the statement that for a 1 - form
$\omega^1 =f_1 dx_1 + f_2 dx_2 + f_3 dx_3$
with domain
$D$
and domain
$\partial D$
,
$\int_D d \omega^1 = \int_{\partial D} \omega^1$
.
The exterior derivative of
$\omega^1$
is
$d \omega^1 =( \frac{\partial f_3}{\partial 2} - \frac{\partial f_2}{\partial 3}) dx_2 \wedge dx_3 +( \frac{\partial f_1}{\partial 3} - \frac{\partial f_1}{\partial 3}) dx_3 \wedge dx_1 + ( \frac{\partial f_2}{\partial 1} - \frac{\partial f_1}{\partial 2}) dx_2 \wedge dx_1$

(1) becomes
$\int_D \frac{\partial f_3}{\partial 2} - \frac{\partial f_2}{\partial 3}) dx_2 \wedge dx_3 +( \frac{\partial f_1}{\partial 3} - \frac{\partial f_1}{\partial 3}) dx_3 \wedge dx_1 + ( \frac{\partial f_2}{\partial 1} - \frac{\partial f_1}{\partial 2}) dx_2 \wedge dx_ = \int_{\partial D} f_1 dx_1 +f_2 dx_2 +f_3 dx_3$
.
which is equivalent to

$\int_D \mathbf{\nabla} \times \mathbf{F} dD = \int_{\partial D} \mathbf{F} \cdot d \mathbf{r}$
where
$\mathbf{F} =(f_1 ,f_2 , f_3)^T$
.