Let $\;F=(F_1,F_2)\;$ be a two-dimensional vector field and
consider the rectangle $\;\mathcal R= PQRS\;$:
If $\;\vec v\;$ is a function which gives outward-facing unit normal
vectors to $\;\partial \mathcal R\;$(=boundary of $\;\mathcal R\;$),
then by divergence theorem one can get:$\;\int_{\mathcal R} div(F_1,F_2) \;dx=\int_{SR} F_2\;dx_1
-\int_{PQ} F_2\;dx_1-\int_{SP} F_1\;dx_2+\int_{QR} F_1\;dx_2\;$
My Attempt:
Searching on google, I found this:
So, in my case it holds:
$\;\int_{\mathcal R} div(F_1,F_2) \;dx=\;\int_{\partial \mathcal R} F_2 dx_1 – F_1 dx_2\;(1)$
In addition, $\;\partial \mathcal R=(SR)\cup(RQ)\cup(QP)\cup(PS)\;(2)$
Now, combining $\;(1),(2)\;$ and considering the counterclockwise direction , I get:
$\;\int_{\partial \mathcal R} -F_2 dx_1 + F_1 dx_2=-\int_{SR} F_2 dx_1+\int_{PQ} F_2 dx_1-\int_{QR} F_2 dx_1+\int_{PS} F_2 dx_1+\int_{SR} F_1 dx_2-\int_{PQ} F_1 dx_2+\int_{QR} F_1 dx_2-\int_{PS} F_1 dx_2\;$
At this point, I've been stuck. Not only I find some additional terms such as $\;\int_{QR} F_2 dx_1\;$, but also some of the signs are wrong.
I haven't seen boundary integrals since a very long time, so I'm $\;100\;$% sure there's something I'm missing here!
Any help would be valuable.
Thanks in advance!
Best Answer
Those additional terms vanish because they are equal to zero. For example, $\int_{QR} F_2\,dx_1$ is an integral along the vertical segment $QR$; since $x_1$ is constant on $QR$, we have that $dx_1=0$ on $QR$, and therefore this whole integral $\int_{QR} F_2\,dx_1=0$.
Also, as @TedShifrin pointed out in comments, your signs in (1) are backwards. Note that the quote from Google shows $\oint_C P\,dy-Q\,dx$. Matching their notation with your notation, you should have substituted $P=F_1$, $Q=F_2$, $y=x_2$ (the "vertical" coordinate), and $x=x_1$ (the "horizontal" coordinate). But you've got them backwards…