TITLE: Show that the form $w$ is closed but not exact
QUESTION [19 upvotes]: Let $$w~=~\dfrac{-y}{x^2+y^2}dx+\dfrac{x}{x^2+y^2}dy, \qquad (x,y)~\in\mathbb{R}^2\backslash \{(0,0)\}.$$
Showing that $w$ is closed is easy. Just calculate $dw$ and you'll get 0.
But how do I show that $w$ is not exact? 
In other words, I need to prove that there is no form $\lambda$ such that $w=d \lambda$
Should I assume that $w=d \lambda$ and try to get to a contradiction?

REPLY [6 votes]: If $w$ is exact, its integral along any contour $C$ should be equal to zero. But, plotting the vector field $\mathbf{E}=(E_x,E_y)$ with
$$E_x=-\frac{y}{x^2+y^2},\qquad E_y=\frac{x}{x^2+y^2},$$
we will get something like this:

and it is clear that the integral $\int \mathbf{E}\cdot d\mathbf{r}$ along the circles centered at the origin does not vanish.