Attempt no 2. Well, it is my first question here, so you may expect some errors, even in grammar, because english is not my mother language. So, I was asked to 'verify' the Stokes Theorem in these questions, and I would like to use differential forms, because it is the content that we are discussing now (and by verify I mean solve both sides of Stokes equation and verify if they are equal):
a) $\omega$ =$(x+3y)dx+(2x-y)dy$ and $\Sigma=\{(x,y)|x^2+2y^2\le2\}$
b) $\omega=xdx+zdy$ and $\Sigma=\{(x,y,z)|x^2+y^2=z^2, 0\le z\le1\}$
c) $\omega=zdxdy$ and $\Sigma=\{(x,y,z)|x^2+y^2+z^2\le 1, z\ge0\}$
I tried to solve the a), watched some video courses, but I simply can't understand how to do it. The only part that I did was $d\omega$ from a), and I got by result $dydx$.
Thanks since now.
Stokes' theorem says $\int_\Sigma d\omega=\int_{\partial\Sigma}\omega.$
For part a), on the left-hand side integrand $d\omega=3dy\wedge dx+2dx\wedge dy=-dx\wedge dy.$ So our integral is
$$ \int_\Sigma d\omega=-\iint_\Sigma dx\,dy=-\int_{-1}^{+1}\int_{-\sqrt{2-2y^2}}^{+\sqrt{2-2y^2}}dx\,dy=-2\int_{-1}^{+1}\sqrt{2-2y^2}\,dy=-\pi\sqrt{2} $$
(Draw a picture to help write these bounds of integration for the region $\Sigma=\{x^2+2y^2\leq 2\}$).
Then to compute the right-hand side $\int_{\partial\Sigma}\omega$ we need a parametrization of the boundary ellipse $\partial\Sigma=\{x^2+2y^2=2\}$, which I will take to be $(\sqrt{2}\cos t,\sin t)$
We have
$$ \int_{\partial\Sigma}\omega = \int(x+3y)dx+(2x-y)dy \\= \int_0^{2\pi} -(\sqrt{2}\cos t+3\sin t)\sqrt{2}\sin t\,dt+(2\sqrt{2}\cos t -\sin t)\cos t\,dt \\=\int_0^{2\pi}(-3\sqrt{2}\sin^2t - 3\sin t\cos t+2\sqrt{2}\cos^2t)\,dt\\=-\pi\sqrt{2}. $$
For that last equality, you can use the identities that $\int_0^{2\pi}\sin t\cos t\,dt=0$ and $\int_0^{2\pi}\cos^2t\,dt=\int_0^{2\pi}\sin^2t\,dt=\pi.$ (If you have trouble remembering these identities, just imagine integrating the Pythagorean identity over one period.)
So Stokes' theorem is verified.