If $\vec F=\frac{\vec r}{r^3}$ then show that $\textrm{div} \vec F = 0$ but $\exists$ is no $\vec G ∶ \mathbb{R}^3 ⧵ {(0,0,0)} \to \mathbb{R}^3$ such that $\vec F = \textrm{curl} \,\vec G$.
I can show that $\textrm{div} \vec F = 0$. I know that if $\textrm{div} \vec\, F = 0$ then there exists a vector $\vec G$ such that $\vec F = \textrm{curl}\, \vec G$.
What is happening here?
Please explain in details why $\vec F \neq \textrm{curl} \,\vec G$ here.
Edit: Consider $\vec F:S\to \mathbb{R}^3$ such that $\vec F=\frac{\vec r}{r^3}$ and $S$ is the unit sphere centred at origin.
Then outward drawn normal $\hat n$ to S is $\hat n=\vec r$ (since $|\vec r|=1$)
Then $$\int\int_S\vec F.\hat n ds=\int\int_S \frac{1}{r}dS=\int\int_SdS=4\pi.$$
But if $\vec F = \textrm{curl}\, \vec G$, then by using Stokes' theorem, we have $$\int\int_S\vec F.\hat n ds=\int\int_S(\textrm{curl} \,\vec G).\hat n ds=\int_C \vec G.d\vec r$$.
Is my way of approach correct? How to find the last integral?