How do I show that a vector field is non conservative?

Question

Given $X \subset \mathbb{R}^2 $ not simply-connected. I want to get an example of a vector field that has zero curl, but is not the gradient of any smooth function on $X$. I was thinking about taking $\left( \frac{-y}{x^2 + y^2} , \frac{x}{x^2 + y^2 } \right) $ assuming without loss of generality there is a hole at 0. But how do I show that there is not such a smooth function f, so that $grad f = \left( \frac{-y}{x^2 + y^2} , \frac{x}{x^2 + y^2 } \right)$ ?

Answer 1

Remember that conservative fields $F$ are path independent, as they can be written as a gradient of a function $f$ (defined on the same domain as $F$): $$ F=\nabla f $$ Consequently, on any curve $C=\{ r(t)\;|\; t\in [a,b]\}$, by the fundamental theorem of calculus $$ \int_C F\, dr = \int_C \nabla f \, dr = f(r(b))-f(r(a)), $$ in other words the integral only depends on $r(b)$ and $r(a)$: it is path independent (it only depends on the endpoints $r(b)$ and $r(a)$, and not on $C$). Another way of writing this statement is: for any pairs of curves $C_1$, $C_2$ having the same endpoints, we should have $$ \int_{C_1} F dr = \int_{C_2} F dr $$

Now, back to your question. Actually, there does exist a function $f$ such that $\nabla f = \left( \frac{-y}{x^2 + y^2} , \frac{x}{x^2 + y^2 } \right)$ (can you find it?).

But in spite of that, the field is not conservative. If it were it should be path independent. But if you compute the integral $\int_C \nabla f \cdot d\vec{r}$ along two different paths having same endpoints, you will get different results (provided you carefully choose those paths)! For example, try the paths from $(1,0)$ to $(-1,0)$, the first one along the circular arc with radius $1$ and $y>0$, the second one with $y<0$.

This will lead you to the following question: how can the field simultaneously not be conservative and be the gradient of a function $f$ ? If you can answer this question you are ready for your exam (the answer to this question is given above :) ).