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?
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.