Start by working directly in $\RR^3$ and using rectangular coordinates. Choose any 1-form $\beta$ in $\RR^3$, expressed in terms of rectangular coordinates. Compute $\alpha=d\beta$. If you get zero, start over; $\beta$ should not be exact. Now switch to spherical coordinates, and evaluate $\alpha$ on the unit sphere, $\SSS^2$. Use what you know!

Now evaluate $\beta$ on the unit sphere, then take $d$ of the result. Did you get the same answer? That is, does exterior differentiation commute with the process of evaluating a differential form on a surface?

The answer should be yes. The result of “evaluating” a differential form on a surface is called the pullback of the form from the larger space to the surface, and pullbacks do commute with $d$.

So what happens if you integrate $\alpha$ over $\SSS^2$? To integrate a differential form, simply drop the wedge products and do the integral — and check whether the orientation agrees with the one you want, inserting a minus sign if it doesn't. Since $\alpha=d\beta$ on the sphere, you can use Stokes' Theorem as discussed in §5.7; your integral must therefore evaluate to zero. Did it?

Let's try this again, working directly on $\SSS^2$, using the standard orientation $\omega=\sin\theta\,d\theta\wedge d\phi$. We know that \begin{equation} \int_{\SSS^2} \omega = 4\pi \ne 0 \end{equation} so that, just as in the previous section, $\omega$ can not be exact. Yet clearly, \begin{equation} \omega = d(-\cos\theta\,d\phi) \end{equation} What is going on here?

The function $\cos\theta$ is well-defined on $\SSS^2$, so the problem must be with $d\phi$. Intuitively, we expect $\sigma^\phi=\sin\theta\,d\phi$ to be well defined; after all, it is a normalized basis 1-form. But \begin{equation} \cos\theta\,d\phi = \cot\theta\,\sigma^\phi \end{equation} and $\cot\theta$ is infinite at the poles. Put differently, $d\phi$ is not defined at the poles, although the combination $\sin\theta\,d\phi$ is reasonably behaved.

Another way to see this is to go back to $\RR^3$ and use rectangular coordinates. We have \begin{equation} x\,dy - y\,dx = r^2\,\sin^2\theta\,d\phi \end{equation} or equivalently \begin{equation} r\,\sin\theta\,d\phi = \cos\phi\,dy-\sin\phi\,dx \end{equation} which shows explicitly that $r\,\sin\theta\,d\phi$ has magnitude $1$ everywhere, including when $x=0=y$ (even though the limit does depend on $\phi$). Multiplying these expressions by $\cot\theta$ clearly leads to a 1-form that is not well behaved, either in $\RR^3$ or on $\SSS^2$, namely $r\,\cos\theta\,d\phi$.