Why in Green's theorem can we not simply integrate $y \mathrm{d} x$ instead of $y \mathrm{d} x - x \mathrm{d} y$? According to the multidimensional Stoke's theorem, in order to evaluate the integral of a form $\omega$, I just have to find a one-form $\alpha$ so that $d\alpha=\omega$ and then use
$\int_{\partial \Omega} \alpha=\int_{\Omega} \omega$
say $\omega$ is just the area form, $dx \wedge dy$. Then, according to Stokes' theorem I just have integrate $-x dy$ around the contour. However, according the Green's theorem I should choose rather the one form $y dx-x dy$. I can see the second one is the right choice but I can't see why the first one is the wrong one. Can anyone tell me the error in my reasoning please?
 A: A version of Stokes' Theorem applicable to your setting states that for a nice compact subset $\Omega \subset \Bbb R^n$ and a nice $(n - 1)$-form $\alpha$ defined on some nice set containing $\Omega$, we have
$$\int_{\partial \Omega} \alpha = \int_\Omega d\alpha ,$$
where the orientation on $\partial \Omega$ is that induced by the orientation on $\Omega$.
Notice that not all integrals $\int_\Omega \beta$ of $n$-forms $\beta$ cannot be evaluated directly using this theorem, but rather only those for which there is some $(n - 1)$-form $\alpha$ such that $d\alpha = \beta$, i.e., only when $\beta$ is exact.
In the special case $n = 2$, any $1$-form can be written as $\alpha = P \,dx + Q\,dy$ for some functions $P, Q$ of $x, y$. Then, $$d\alpha = \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx \wedge dy .$$ (Making the usual identifications and requiring that $\partial \Omega$ but oriented anticlockwise then immediately gives the classical statement of Green's Theorem in terms of the vector field $\alpha^\sharp = P \partial_x + Q \partial_y$.)
So for any $1$-form $\alpha$ such that $d\alpha = dx \wedge dy$---equivalently, for any nice functions $P, Q$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$---we have
$$\operatorname{area}(\Omega) = \int_\Omega dx \wedge dy = \int_\Omega d\alpha = \int_{\partial \Omega} \alpha ,$$
and there are infinitely many such $1$-forms. Indeed, if we take any $1$-form whose exterior derivative is $dx \wedge dy$, say, $$\alpha_0 := \frac{1}{2} \left(-y \,dx + x \,dy\right), $$ then for any nice function $f$ on $\Omega$, we again have
$$\require{cancel}d(\alpha_0 + df) = d\alpha_0 + \cancelto{0}{d^2\! f} = dx \wedge dy .$$
Indeed, taking $f(x, y) = \frac{1}{2} x y$ gives $df = \frac{1}{2} (y\,dx + x\,dy)$ and $\alpha_0 + df = x\,dy$, and hence
$$\operatorname{area}(\Omega) = \int_{\partial\Omega} x\,dy .$$
