## Green's Theorem and the Flux Out of a Surface

Suppose we have a force field
$\mathbf{F} =(f_1 , f_2 )$
acting on a surface
$B$
in the
$xy$
plane bounded by a curve
$C$
. We can parametrize the curve
$C$
in terms of the distance
$s$
along it, then a normal is
$\mathbf{n} = ( \frac{dy}{ds} , - \frac{dx}{ds})$

Then
$\oint_C \mathbf{F} \cdot \mathbf{n} ds = \oint_C f_1 \: dy -f_2 \: dx= \oint_C f_1 \: dy + (-f_2 ) \: dx$

According to Green's Theorem
$\oint_C P \:dx + Q \: dy = \int \int_R ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} ) dx \: dy$
. Put
$P=-f_2 , \: Q=f_1$
then
$\oint_C -f_2 \:dx + f_1 \: dy = \int \int_B ( \frac{\partial f_1}{\partial x} + \frac{\partial f_2}{\partial y} ) dx \: dy$

which we can write
$\oint \mathbf{F} \times d \mathbf{r} = \int \int div \cdot \mathbf{F} \: dx \: dy$

The hand hand side is the flux of
$\mathbf{F}$
out of the surface.