Prove: The pullback of a volume form on a sphere to a cylinder is a volume form
We denote $S = \{ (x,y,z) \mid x^2 + y^2 + z^2 = 1\}$,$ C = \{ (x,y,z) \mid x^2 + y^2 = 1 , |z| < 1 \}$.
Given a volume form $\mu$ on $S$, prove that $\phi^*\mu$ is a volume form on $C$, where $\phi(x,y,z) =(x\sqrt{(1-z^2)},y\sqrt{(1-z^2)},z). $
Is it not true that $\phi^*\mu = \mu(\phi(x),D\phi_1,D\phi_2,D\phi_3)$, where $D\phi_i$ is the i'th column of the differential of $\phi$ at point $x$?
To finish the proof, I need to show that the form never vanishes on the cylinder (is this correct?). To do this, I'm pretty sure I should do something with the basis of the tangent space of the sphere, which is just the standard basis.
I'm not really sure how to show that it doesn't vanish though, and any help would be appreciated!
Because you're dealing with surfaces (rather than solids) you're speaking of area forms. (It's not inconsistent to call them "volume" forms as long as you understand "$2$-dimensional volume", bit this means your expression for $\phi^{*}\mu$ is not quite right.)
Cylindrical coordinates are convenient for describing both surfaces, and for expressing the horizontal projection mapping $\phi$. Your easiest approach is probably to:
Parametrize the unit cylinder $C$ using $(\theta, z)$,
Express the mapping $\phi:C \to S$ in terms of the parameters of $C$, i.e., pick a suitable coordinate system $(u, v)$ on the sphere and write $(u, v) = \phi(\theta, z)$,
Show the mapping $\phi$ is regular, i.e., $\det D\phi$ is non-vanishing. (If you need the actual area forms (which encode a famous theorem of Archimedes), write the area form of $S$ in terms of $(u, v)$ and express it in terms of $(\theta, z)$, i.e., compute the pullback the area form of $S$.)