Q:

Classic example of a non exact form

Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$. 
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.

A:

I'm not sure whether this is acceptable or not, but the simplest thing to do (since the form is closed) is to attempt to integrate it to find supposed $\phi$ such that your form $\omega$ is $\mathrm d \phi$.
One can derive a contradiction most straightforwardly by going around a circle, say $x^2+y^2=1$. We use the path $x(t)=\cos t,y(t)=\sin t$.
$$\int_0^{2\pi} \omega = \int_0^{2\pi} \frac{\cos^2 t +\sin^2 t}{1}\mathrm d t = 2\pi$$
But if $\omega=\mathrm d \phi$ then the expression on the left is $\phi((1,0))-\phi((1,0))=0$, a contradiction.

In response to your complaints about this method: The definition of exact is that $\omega=\mathrm d \phi$ for some $\phi$. The only result we've used is that integrals of derivatives are given by boundary values (a very mini version of Stokes's theorem), i.e. 
$$\int_{\mathbf x}^{\mathbf y} \mathrm d \phi = \left[\int_{t_x}^{t_y} \frac{\mathrm d \phi}{\mathrm d t} \mathrm d t = \right]\phi({\mathbf y})-\phi({\mathbf x})$$
which I don't think is really a mysterious result. If there's something you're not happy with do just say.

