if we have $g(\phi,\theta)=\begin{bmatrix}r\cos\theta\\r\sin\theta\\\sqrt{1-r^2} \end{bmatrix}$ and we have $\omega=zdx \wedge dy \in A^2(R^3)$ i want to find the expression for $g^*\omega$
I know that $g^*\omega = g\cdot f$ my textbook says that it is $\sqrt{1-r^2}*rdr\wedge d\theta$. Okay, I know where the SQRT expression came from, it came from z. But what about the rest? how was $dx\wedge dy$ converted to $r dr\wedge d\theta$??
You have a 2-form $\omega:=zdx\wedge dy\in\Omega^2(\mathbb R^3)$ and a map $g$ such that $g(\phi,\theta)=(r\cos\theta,r\sin\theta,\sqrt{1-r^2})$ and we have to compute the pullback $g^*\omega$ (the coordinate system $\{x,y,z\}$ of $\mathbb R^3$ is global). $$\begin{aligned}g^*\omega & =\omega\circ g=\omega(g(\phi,\theta))\\&=z\circ g(\phi,\theta)d[x\circ g(\phi,\theta)]\wedge d[y\circ g(\phi,\theta)]\\ &=\sqrt{1-r^2}d[r\cos\theta]\wedge d[r\sin\theta]\\ &=\sqrt{1-r^2}[\cos\theta dr-r\sin\theta d\theta]\wedge[\sin\theta dr+r\cos\theta d\theta]\\ &=\sqrt{1-r^2}(\underbrace{\cos\theta\sin\theta dr\wedge dr}_{=0}+r\cos^2\theta dr\wedge d\theta\underbrace{-r\sin^2\theta d\theta\wedge dr}_{=r\sin^2\theta dr\wedge d\theta}-\underbrace{r^2\sin\theta\cos\theta d\theta\wedge d\theta}_{=0})\\ &=\sqrt{1-r^2}r(\cos^2\theta+\sin^2\theta)dr\wedge d\theta=\boxed{r\sqrt{1-r^2}dr\wedge d\theta}. \end{aligned}$$ Remember that $ x,y,z:\mathbb R^3\to\mathbb R$ acts on point of the real space as projections.