I've been given the following differential form:
$$\eta = -\frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy$$
and need to express it in the polar coordinates $(x,y) = (r\cos\theta, r\sin\theta)$. To do so I did the following calculation:
$$-\frac{y}{x^2+y^2} = -\frac{r\sin\theta}{r^2} = -\frac{\sin\theta}{r}$$ $$\frac{x}{x^2+y^2} = \frac{r\cos\theta}{r^2} =\frac{\cos\theta}{r}$$
and
$$dx = d(x(r\cos\theta, r\sin\theta)) = \frac{\partial(r\cos\theta)}{\partial r}dr+\frac{\partial(r\cos\theta)}{\partial \theta}d\theta = \cos\theta dr -r\sin\theta d\theta$$ $$dy = d(y(r\cos\theta, r\sin\theta)) = \frac{\partial(r\sin\theta)}{\partial r}dr+\frac{\partial(r\sin\theta)}{\partial \theta}d\theta = \sin\theta dr +r\cos\theta d\theta$$
So the final answer should be $$(\sin^2\theta+\cos^2\theta) \,d\theta+\frac{(\sin\theta\cos\theta-\sin\theta\cos\theta)}{r}\,dr= d\theta$$
Is this correct? More generally, if a form is $$\omega = \omega_xdx + \omega_ydy$$ then does the application of some pullback $\Phi$ yield
$$\Phi^{*}\omega =(\omega_x\circ\Phi)(d\Phi^{*}x)+(\omega_y\circ\Phi)(d\Phi^{*}y)$$
always?