Why is $\int_{\partial D}x\,dy$ invalid for calculating area of $D$? I am just learning about differential forms, and I had a question about employing Green's theorem to calculate area. Generalized Stokes' theorem says that $\int_{\partial D}\omega=\int_D d\omega$. Let's say $D$ is a region in $\mathbb{R}^2$. The familiar formula to calculate area is $\iint_D 1 dxdy = \frac{1}{2}\int_{\partial D}x\,dy - y\,dx$, and indeed, $d(x\,dy - y\,dx)=2\,dx\,dy$. But why aren't we allowed to simply use $\int_{\partial D}x\,dy$? Doesn't $d(x\,dy)=d(x)dy = (1\,dx + 0\,dy)dy = dx\,dy$?
 A: You can use $\int_{\partial D} x\,dy$ to compute area in this context. The "familiar formula" does have a more symmetric look to it -- maybe that's why you find it more familiar. 
There are infinitely many formulas like this that work. In general you need two functions $P$ and $Q$ such that $Q_x-P_y=1$. Then $\int_{\partial D} P\,dx+Q\,dy$ will compute the area.
$P=-y/2$ and $Q=x/2$ gives your familiar formula.
$P=0$ and $Q=x$ is the formula in question.
One could also use $P=-y$ and $Q=0$ (i.e. $\int_{\partial D} -y\,dx$) to compute the area. 
Those 3 choices are standard ones presented in traditional multivariate calculus texts. But of course there are infinitely many other choices as well.
A: I would like to point out the integrating $xdy$ to get area has a natural geometric interpretation:  you are summing the areas of small horizontal rectangles.  The sign of these areas is determined by whether you are moving up or down, and the sign of x.  Draw a picture of a wild blob, intersect it with a horizontal line, and see how at each intersection point you will get a rectangle.  The areas of these rectangles will cancel out when you are not inside the blob.
Here is my blob:

if I am integrating $xdy$ around this closed curve in the indicated direction, then I am going to move incrementally along the curve a little step at a time, and keep a running sum of $x$ times the small vertical distance I just traveled.  Moving up is a positive distance, moving down is a negative distance.  That is just the signed area of a little rectangle whose length is my current position to the $y$ -axis, and whose height is the small height I just traversed.
I will denote positive areas as green, and negative areas as red.

This point is positive since I am moving up, and $x$ is positive. Some time later I come back to the same height, but at a different $x$ value.

This time the area is negative because I am moving down, and $x$ is positive.
Later on I am again at the same $y$ value:

moving up, x pos

moving down, x negative!
Putting these all together, I have something like this:

Now the red and part of the first green rectangle cancel, and you can see that I only have the area of the horizontal strip which is inside the blob:  the orientations have automatically recorded the difference between inside and outside!

Hopefully this application of Green's theorem is not so mysterious any more.
