differential 1 - form of $\omega$ Let $\omega$ be a differential form.
$$\omega = \dfrac{xdy\wedge dz + ydz\wedge dx + zdx\wedge dy}{(x^2 + y^2+ z^2)^{3/2}}$$
1)Find $d\omega$.
2)show that exist 1-form $\theta$ such that $d\theta = \omega$ in half-space $z>0$.
3) Show that doesn't exist 1-form $\theta$ such that $d\theta = \omega$ in $ \mathbb R^3 \setminus\{0\} $
I found $d\omega$, it  equals zero, but i have no idea how to show in 2) and 3). Can you help me with this?
 A: 2) You have shown in the first question that $\omega$ is closed, whence using Poincaré's lemma, it is exact on each contractible subsets of $\mathbb{R}^3$ on which $\omega$ is defined, such that the half-space $z>0$, since it is convex.
3) If $\omega$ were exact on $\mathbb{R}^3\setminus\{0\}$, using Stokes' theorem, its integral on $S^2$ would be zero, since $S^2$ has no boundary, but it is equal to $4\pi$.
I feel a bit guilty about not computing this integral, so let's do it:
$$\begin{align}\int_{S^2}\omega&=\int_{S^2}x\,\mathrm{d}y\wedge\mathrm{d}z+y\,\mathrm{d}z\wedge\mathrm{d}x+z\,\mathrm{d}x\wedge \mathrm{d}y,\tag{1}\\\tag{2}&=\int_{B^3}\mathrm{d}(x\,\mathrm{d}y\wedge\mathrm{d}z+y\,\mathrm{d}z\wedge\mathrm{d}x+z\,\mathrm{d}x\wedge\mathrm{d}y),\\\tag{3}&=3\int_{B^3}\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z,\\&=4\pi.\tag{4}\end{align}$$
Here are the steps behing each equality:


*

*$(1)$: $\omega_{\vert S^2}=x\,\mathrm{d}y\wedge\mathrm{d}z+y\,\mathrm{d}z\wedge\mathrm{d}x+z\,\mathrm{d}x\wedge\mathrm{d}y$, since $x^2+y^2+z^2=1$ for $(x,y,z)\in S^2$,

*$(2)$: Stokes' theorem on $B^3$, noticing that $\partial B^3=S^2$,

*$(3)$: $\mathrm{d}(x\,\mathrm{d}y\wedge\mathrm{d}z+y\,\mathrm{d}z\wedge\mathrm{d}x+z\,\mathrm{d}x\wedge\mathrm{d}y)=3\,\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$, using standard properties of the wedge product and the exterior derivative,

*$(4)$: $\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$ is the Lebesgue measure on $\mathbb{R}^3$ and the Lebesgue measure of $B^3$ is $4\pi/3$.


I hope that everything is crystal clear.
