TITLE: Is there a vector field that is the complete opposite of a conservative one
QUESTION [6 upvotes]: Is there a three-dimensional vector field such that for every non-selfintersecting closed curve (that is not just one point, to avoid degenerate cases) the respective line-integral on the curve becomes non-zero?
If not, what if I know that every point of the curve has all its coordinates positive?

REPLY [7 votes]: It cannot be done, not even locally: Consider the unit circle $\gamma_0$ in the $(x,y)$-plane made out of wire and oriented counterclockwise. Let $$\oint_{\textstyle\gamma_0 }{\rm F}\cdot d{\bf x}=:c\ne0\ .$$ Now turn the circle in space a total of $180$ degrees, using the $x$-axis as axis, with intermediate positions $\gamma_t$ $\>(0\leq t\leq\pi)$. At time $\pi$  we are again at the starting position, but with the sense of direction reversed. Then
$$\oint_{\textstyle\gamma_\pi}{\rm F}\cdot d{\bf x}=-c\ .$$
By continuity there has to be a $t\in\ ]0,\pi[\ $ for which
$$\oint_{\textstyle\gamma_t}{\rm F}\cdot d{\bf x}=0\ .$$