How do I show that $F^{∗}(dx∧dy∧dz) = ρ^{2} \sin φ dρ∧dφ∧dθ$.

Question

I dont know how to solve. Please help me. I need to understand such types of the question for my exam studyings.

Answer 1

Since $x=\rho\sin\phi\cos\theta,y=\rho\sin\phi\sin\theta,z=\rho\cos\phi$, you can get \begin{eqnarray*} dx&=&\sin\phi\cos\theta d\rho+\rho \cos\phi\cos\theta d\phi-\rho\sin\phi\sin\theta d\theta, \\ dy&=&\sin\phi\sin\theta d\rho+\rho \cos\phi\sin\theta d\phi+\rho\sin\phi\cos\theta d\theta, \\ dz&=&\cos\phi d\rho-\rho\sin\phi d\phi. \end{eqnarray*} Using $d\rho\land\rho=0, d\phi\land d\phi=0, d\theta\land d\theta=0, d\rho\land d\phi=-d\phi\land d\rho, d\rho\land d\theta=-d\theta\land d\rho, d\theta\land d\phi=-d\phi\land d\theta$, it is not hard to get \begin{eqnarray*} F^*(dx\land dy\land dz)&=&(\sin\phi\cos\theta d\rho+\rho \cos\phi\cos\theta d\phi-\rho\sin\phi\sin\theta d\theta)\\ &&\land(\sin\phi\sin\theta d\rho+\rho \cos\phi\sin\theta d\phi+\rho\sin\phi\cos\theta d\theta)\land(\cos\phi d\rho-\rho\sin\phi d\phi)\\ &=&\rho^2\sin\phi d\rho\land d\phi\land d\theta. \end{eqnarray*}