Let $f: \Bbb{R}^2 \rightarrow \Bbb{R}^3$
be given the $2$-form $\omega = x \, dy \wedge dz$ compute the pull back if $f$ is defined via
$$(\theta, \phi) \mapsto ((2+\cos \phi)\cos \theta, (2+\cos \phi)\sin \theta, \sin \phi)$$
Then I get
\begin{align} f^*\omega&= (2+ \cos \phi)\cos \theta \,d((2+\cos \phi)\sin \theta) \, \wedge d(\sin \phi)\\ &= (2+ \cos \phi)\cos \theta (-\sin \phi \sin \theta \,d \phi+(2+\cos \phi)\cos \theta \, d \theta) \, \wedge \cos \phi d \phi\\ &=(2+\cos \phi)^2\cos^2 \theta \cos \phi \, d \theta \, \wedge d \phi \end{align}
My question is does the $d \phi \wedge d \phi$term cancel? After distributing the wedge product. If so then /I can take it from here.