# Sufficient condition for conservative fields of class $\mathcal C^1$

I would need help for a complete and detailed demonstration of this theorem if it is possible. I found several handouts on the net but I wondered if there was a demonstration not too long and complicated.

Theorem: Let $$D\subseteq \mathbb R^3$$ a non-empty open and $$\overline F:D\rightarrow \mathbb R^3$$ a vector field of class $$\mathcal C^1$$ with $$\overline F=(F_x,F_y, F_z)=f(r)\widehat{\bf r}$$ where:

$$F_x=f(r) \frac xr, \quad etc.$$

If $$D$$ is a domain simply connected - the differential form is therefore exact - and for every $$\overline r=(x,y,z)\in D$$ are valid the symmetry conditions

$$$$\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x}, \quad \frac{\partial F_x}{\partial z}=\frac{\partial F_z}{\partial x}, \quad \frac{\partial F_y}{\partial z}=\frac{\partial F_z}{\partial y}$$$$

hence the vectorial function $$\overline F$$ is conservative. In addition $$\overline \nabla \times F=0$$, $$\forall\,\, \overline r\ne 0$$.

• Isn't the mathematics section of SE a better place to ask this pure mathematics question? – DanielC Dec 8 at 23:50
• @DanielC With so much honesty, I didn't get along so well. Free offenses of all kinds, and afterwards I was wiped out of the community of mathematicians. I feel great among LaTeX users and here. – Sebastiano Dec 8 at 23:57