## The Divergence Theorem in Terms of Forms

We can write the Divergence Theorem in terms of forms. In fact, if
$\omega^2 = f_1 dx_2 \wedge dx_3 + f_2 dx_3 \wedge dx_1 +f_3 dx_1 \wedge dx_2$
is defined on a region
$D$
of
$\mathbb{R}^3$
with surface
$S$
then the statement
$\int_D d \omega^2 = \int_S \omega^2$
(1) is equivalent to the Divergence Theorem.
To see this not that with
$\omega^2$
as above,
$d \omega^2 = (\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3})dx_1 \wedge dx_2 \wedge dx_3$

$\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3}$
is the divergence of the vector field
$\mathbf{F} = (f_1,f_2,f_3)^T$

Substituting for
$\omega^2 , \: d \omega^2$
in (12) gives
$\int_D (\frac{\partial f_1}{\partial x_1} + \frac{\partial f_2}{\partial x_2} + \frac{\partial f_3}{\partial x_3})dx_1 \wedge dx_2 \wedge dx_3 = \int_S f_1 dx_2 \wedge dx_3 + f_2 dx_3 \wedge dx_1 +f_3 dx_1 \wedge dx_2$

This is equivalent to the Divergence Theorem.