Find the Poincare Dual of a ray in $\mathbb{R}^2-\{0\}$

Solution 1:

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.