Solving problem from textbook, not homework.
Define the $r\theta$-plane on $\mathbb{R}^2$. Define 0-forms $f=r \cos \theta$ and $g = r \sin \theta \in \Omega^0(\mathbb{R}^2)$. Problem: compute $df$, $dg$ and $df \wedge dg$ w.r.t. $dr$ and $d \theta$.
My answer. For 0-forms the exterior derivative is defined as $$df = \sum \frac{\partial f_I}{\partial x_I} dx_I,$$ and the wedge product defined as $$\omega \wedge \tau = \sum_{I,J} (a_I b_J) dx_I \wedge dx_J.$$ Computing the exterior derivatives \begin{align} df &= \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta}d\theta &= \cos \theta dr - r \sin \theta d\theta \end{align} and \begin{align} dg &= \frac{\partial g}{\partial r}dr + \frac{\partial g}{\partial \theta}d\theta &= \sin \theta dr + r \cos \theta d\theta. \end{align}
Computing the wedge product \begin{align} df \wedge dg &= (\cos \theta \sin \theta)dr \wedge d\theta - (r^2 \sin \theta \cos \theta)d\theta \wedge dr. \end{align}
Is this correct?