Let $\omega = \alpha \,\mathrm{d}r+r\beta\,\mathrm{d}\theta$ a 1-form on $\mathbb{R}^2$, where $\alpha,\beta$ are pulled back by polar coordinates.
I want to find equations for $\alpha,\beta$ where $\mathrm{d}\omega =0$.
My attempt:
$\mathrm{d}\omega = \mathrm{d}(\alpha \mathrm{d}r) + \mathrm{d}(r\beta \mathrm{d}\theta)\\= \mathrm{d}\alpha \wedge\mathrm{d}r + \mathrm{d}(r\beta) \wedge\mathrm{d}\theta=\left (\displaystyle\frac{\partial \alpha}{\partial r}\mathrm{d}r + \displaystyle\frac{\partial \alpha}{\partial \theta}\mathrm{d}\theta\right ) \wedge\mathrm{d}r + \left (\mathrm{d}r\beta + r\mathrm{d}\beta \right )\wedge \mathrm{d}\theta =\displaystyle \frac{\partial \alpha}{\partial \theta}\mathrm{d}\theta \wedge \mathrm{d}r + \beta\mathrm{d}r\wedge \mathrm{d}\theta+r\displaystyle \frac{\partial \beta }{\partial r}\mathrm{d}r\wedge\mathrm{d}\theta = \left (\beta +r\displaystyle \frac{\partial \beta}{\partial r} - \displaystyle \frac{\partial \alpha}{\partial \theta} \right )(\mathrm{d}r\wedge\mathrm{d}\theta)$
So the smooth coefficient of the 2-form has to vanish. I expected something more concise, since this is from a book.