I am currently practicing integration on Manifolds for an exam, but I feel like I haven't fully grasped the concepts 100% yet. So the question reads:
Integrate the 2-form:$$\omega = (6z^2-sin(y))dy\wedge dz + dx\wedge dy$$ over a torus parametrized by: $$T = (\cos(\theta)(R+r\cos(\varphi), \sin(\theta)(R+r\cos(\varphi), r\sin(\varphi))$$Using Stokes Theorem. $\\$ What is the Differential form in polar coordinates?
Now my understanding (and also having looked through some of the answers on Stack Exchange) is that integrating over a torus in $\mathbb{R}^3$ is the same as integrating over a 2-sphere. So I got that: $$\partial\omega = (12z)dz\wedge dy\wedge dz - \cos(y)\wedge dy\wedge dy\wedge dz + dx\wedge dy = dx\wedge dy$$ Then I tried to integrate this over the sphere in $\mathbb{R}^3$, which yielded: $$\int_{T}\omega = \int_{S^2}\omega$$ Using Stokes' Theorem, I got: $$\int_{S^2}\omega = \int_{B^3}\partial \omega = \int_{B^3}dx\wedge dy = \pi r^2$$ Considering that the ball is the boundary of the sphere. I feel like I'm missing something or I'm just not getting the fundamentals of the whole integration on Manifolds idea. If anyone could help me out it would be greatly appreciated!
You have a few mistakes. Here is the correct reasoning.
Stokes' theorem says that $\int_{\partial M} \omega = \int_M d\omega$, where $M$ is a compact oriented manifold with boundary $\partial M$. In your case, $\partial M$ is the torus $T^2$ and $M$ is the solid "donut" which has the surface of the torus as its boundary.
Now in your case, $d \omega = 0$! So $\int_{T_2} \omega = \int_{\partial M} \omega = \int_{M} d \omega = \int_M 0 = 0$.
(To point out a computational mistake: $d(dx\wedge dy) = 0$, not $dx \wedge dy$ as you have written.)