TITLE: Exterior derivative of a complicated differential form
QUESTION [5 upvotes]: Let $\omega$ be a $2$-form on $\mathbb{R}^3\setminus\{0\}$ defined by
  $$ \omega = \frac{x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge
> dy}{(x^2+y^2+z^2)^{\frac{3}{2}}} $$  Show that $\omega$ is closed but
  not exact.

What I have tried so far: In order to show that $\omega$ is closed, I need to show that $d\omega=0$. I'm having some problems getting all of the calculus right and somewhere along the way I'm messing up. I started by rewriting $\omega$ as
$$
\omega = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}}
$$
Now I should be able to use the product rule to evaluate (I think). Then
$$
d\omega = (dx\wedge dy\wedge dz+dy\wedge dz\wedge dx +dz\wedge dx\wedge dy)(x^2+y^2+z^2)^{-\frac{3}{2}} + (\ast)
$$
where 
$$
(\ast) = (x\,dy\wedge dz+y\,dz\wedge dx +z\,dx\wedge dy)\left(-\frac{3}{2}(2x\,dx+2y\,dy+2z\,dz)\right)(x^2+y^2+z^2)^{-\frac{5}{2}}
$$
Even after trying to simplify everything, I can't get it to cancel. This makes me think that perhaps I can't apply the product rule like this. 
What should I do to calculate $d\omega$?
If $\omega$ is a globally defined smooth form and if $d\omega=0$, then $\omega$ is exact because there is some other form $\alpha$ with $d\alpha=\omega$ and $d^2\alpha=d\omega=0$. Because $\omega$ is not defined at $(0,0,0)$, it makes sense that it isn't exact. 
Is there a way to use the above reasoning to show that there can't be an $\alpha$ such that $$d\alpha=\omega?$$

REPLY [3 votes]: Use spherical coordinates. In spherical coordinates $r,\theta,\phi$, the form reads:
$$ \omega = \sin\theta\,d\theta\wedge d\phi. $$
This is closed, because the coefficient only depends on a coordinate that is already used:
$$ d\omega = d(\sin\theta)\wedge d\theta\wedge d\phi = \cos\theta\,d\theta\wedge d\theta\wedge d\phi = 0, $$
since $d\theta\wedge d\theta=0$.
This is anyway not exact, because you can integrate it on the sphere!
Integrating it on the closed area $0\le\theta\le \pi, 0\le\phi\le2\pi$, that is a sphere (of radius one), you find (it's a simple calculation):
$$\oint\omega = 4\pi.$$
Therefore, the form is not exact.