Find the Poincare Dual of a ray in $\mathbb{R}^2-\{0\}$ This is example-exercise 5.16 in Bott and Tu (which I'm independently reading through.)  The problem states: Let $M=\mathbb{R}^2-\{0\}$, and $X\subseteq M$ be the closed submanifold $\{(x,0):x>0\}$.  Show that the Poincare dual of $X$ is $d\theta/2\pi$.  
My attempt at a solution:  We want to show that for all $\omega\in H_c^1(M)$, $\int_X\omega=\int_{M}\omega\wedge d\theta/2\pi$.  So let $\omega=fdr+gd\theta$ be a closed 1-form on $M$ with compact support.  It follows that $f$ and $g$ have compact support, and $d\omega=0$ so $\partial f/\partial \theta=\partial g/\partial r$ everywhere on $M$.  
We want to use this to show that $\int_X\omega=\int_{0}^\infty f(r,0)dr=\frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^\infty f(r,\theta)drd\theta$.  So if for some reason $f$ doesn't depend on $\theta$ then we are done.  But I don't see how to show this, much less how to use the closed assumption any further.  Any hints/suggestions/errors in my work? 
 A: There are fancier ways to do this, which you'll learn eventually: If a compact group $G$ acts on $M$, then any closed $k$-form $\omega$ on $M$ is cohomologous to a $G$-invariant closed $k$-form $\tilde\omega$ on $M$ (i.e., $\omega-\tilde\omega = d\eta$ for some $k-1$-form $\eta$). 
But let's just do this bare-hands here. Consider 
$$\frac d{d\theta}\int_0^\infty f(r,\theta)dr = \int_0^\infty \frac{\partial f}{\partial \theta}dr = \int_0^\infty \frac{\partial g}{\partial r}dr = 0,$$
since $g$ has compact support in $M$. So, evidently,
$$\int_0^\infty f(r,\theta)dr = \int_0^\infty f(r,0)dr \text{ for all } \theta\in [0,2\pi],$$
and your result follows.
A: Another way to see why the integral along a ray doesn't depend on the inclination of the ray is to see that, by Stokes, picking finite parts of the rays with long enough length as to contain the support of the form, the difference of the integrals on those rays will be equal to $0$, since the integral on the "circular" boundary parts is zero (the form is $0$ there) and the form $\omega$ is closed.
