Let $\omega=\sin\varphi\,\mathrm{d}\theta\wedge\mathrm{d}\varphi$ be a $2$-form on $\mathbb{R}^3\setminus\{0\}$. Then
$$\int_{\mathbb{S}^2}\omega=\int_{\mathbb{S}^2}\sin\varphi\,\mathrm{d}\theta\,\mathrm{d}\varphi=4\pi\ne0,$$
so $\omega$ is not exact. On the other hand,
$$\omega=\mathrm{d}\eta\text{, where }\eta=\cos\varphi\,\mathrm{d}\theta,$$
which contradicts the fact that $\omega$ is not exact. How is this possible?
Best Answer
There are several things related, let me summarize a little bit.
First of all, the one form $\cos \varphi d\theta$ is not defined on $\mathbb R^3\setminus\{0\}$, but only on $\mathbb R^3\setminus \{x=y=0\}$. That's because polar coordinate is generally not defined at $\{x = y =0\}$.
You may, of course, have the same objection to $\omega = \sin\varphi d\theta \wedge d\varphi$. However, this one is actually defined on $\mathbb R^3\setminus\{0\}$ as it can also be written as
$$\omega(X, Y) = \det\left(X\ \ Y\ \ \frac{\vec r}{r}\right).$$
where we think of $X, Y$ as column vectors, $\vec r$ is the position vector and $r = |\vec r|$. (similar calculations can be found in Is this differential 2-form closed)
We again look into your argument. It is true that $\omega$ is not exact, and you have just shown that $\omega$ is exact when restricted to (e.g.) $\mathbb S^2\setminus \{\text{North pole, South pole}\}$. This make sense, as $\mathbb S^2$ minus two points is homotopic equivalent to $\mathbb S^1$, and so all two forms on $\mathbb S^2$ minus two points must be exact.