Q:

Volume form on $S^2$

I have some basic understanding problem on this. Any help is appreciated:
The volume form of $S^2 \subset R^3$ is given by
$$ \omega = x \ dy \land dz-y \ dx \land dz +z \ dx \land dy$$
In polar coordinates this becomes
$$ \omega = sin\Theta\ d\Theta \land d\phi$$
A volume form ought to be non-vanishing everywhere on the sphere. But how can this be since at the pole $\Theta=0$ the form seems to vanish identically?
Second confusion: A volume form cannot be exact. If it was, by the use of Stokes law, the area of the sphere would be identically to zero which cannot be true.
However if one writes, 
$$ \omega = sin\Theta\ d\Theta \land d\phi=d(-cos\Theta \land d\phi)\equiv d\Lambda$$
where $\Lambda=-cos\Theta \ d\phi$ $,\omega$ seems closed!?. $\Lambda$ appears single-valued,smooth and everywhere well defined on the sphere. What is wrong with $\Lambda$? How can one see that $\omega$ is not exact without resorting to the argument based on Stokes law.
Many thanks!

A:

I just came across this illuminating discussion at StackExchange which answers part of the questions:
Is $\omega=\sin\varphi\,\mathrm{d}\theta\wedge\mathrm{d}\varphi$ an exact form?

