question: Does there exist smooth vector field $ \bf{F}$ $=(F^{1},F^{2},F^{3})$ in $\mathbb{R}^{3}$ such that it is not constant and that integral from this field $\bf{F}$ along any closed curve and every closed surface equals 0?
what I alreay did? I know that the conditions put on value of integral implies: $curl(\bf{F}$ $) = 0$ , $div(\bf{F}$ $)=0$, giving conditions:
$ F_{y}^{3}=F_{z}^{2} \\ F_{z}^{1}=F_{x}^{3} \\ F_{x}^{2}=F_{y}^{1} \\ F_{x}^{1}+F_{y}^{2}+F_{z}^{3}=0 $
Checked that $\bf{F}$ $=(x,y,-2z)$ holds the above conditions.
Can any state whether my reasoning is correct? Thanks in advance.