Question about non conservative vector field

Question

As I know if $$\partial f/\partial y=\partial f/\partial x$$ then f(x,y) is conservative, but there are two counterexamples $$\vec F=\frac{-y\hat\imath+x\hat\jmath}{x^2+y^2}$$ and $$\vec F=\hat\jmath$$

I know there is an similiar question easy question about conservative vector fields

But I still don't understand these two counterexamples contradict with $\partial f/\partial y=\partial f/\partial x$

Can anyone show me how these two counterexamples are non conversative?

thanks

They don't speak at all... so they are not conversative – Emanuele Paolini Feb 07 '15 at 08:26 — Emanuele Paolini, Feb 07 '15 at 08:26

score 1 · Accepted Answer · answered Feb 07 '15 at 08:35

A vector field $\vec F$ is conservative if there exist a function (called potential) $g$ such that $\vec F = \nabla g$. Now it comes out that the integral of $\nabla g$ on any curve $\gamma$ is easily computed as the difference of $g$ on the endpoints of $\gamma$ (this is the fundamental theorem of calculus). So the integral of a conservative vector field along a closed curve is always zero. The converse can be proven to be true, also.

So to prove that a vectorfield $\vec F$ is not conservative it is enough to find a closed curve $\gamma$ on which the integral of $\vec F$ is not zero. In the case of your first example such a curve is any circle centered in the origin.

Your second example $\vec F = j$ is instead an example of a conservative field, since $\vec F = \nabla g$ with $g(x,y,z) = y$.

My professor told me that $\vec F =j$ is non conservative – Simple Feb 07 '15 at 20:32 — Simple, Feb 07 '15 at 20:32

Question about non conservative vector field

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