Changing a Differential Form into Spherical Polars

Question

I am currently trying to integrate:$$\omega_{a,b,c} = \frac{(x-a)dy\wedge dz+(y-b)dz\wedge dx+(z-c)dx\wedge dy}{[(x-a)^2+(y-b)^2+(z-c)^2]^{3/2}}$$ over the unit sphere centered at $(a,b,c)$, with the standard orientation. I am supposed to show that this integral gives $-4\pi$. I am confused by how the spherical coord system of this differential form would look. I know that the form replacing $(x-a)$ with $x$, same with $y$ and $z$ gives: $$\sin\phi\, d\theta\wedge d\phi$$Am I allowed to use the same parametrization? Or would this give a completely different parametrization? I am looking for a way without using Stokes' Theorem, as this is required for another part of the question. Thank you.

Answer 1

Start by considering $(a,b,c) = (0,0,0)$. Set $D = [0,2\pi] \times [0,\pi]$ and let $\Phi \colon D \rightarrow S^2$ be the map given by

$$ \Phi(\theta, \varphi) = (\cos (\theta) \sin(\phi), \sin(\theta) \sin(\phi), \cos(\phi)) $$

(so $x = \cos(\theta) \sin(\phi)$, etc). Since $\Phi|_{D^{\circ}}$ is an orientation preserving diffeomorphism from $D^{\circ}$ onto a subset of $S^2$ whose complement is of measure zero, we have

$$ \int_{S^2} \omega = \int_D \Phi^{*}(\omega) $$

so it is enough to calculate the pullback of $\omega$ along the spherical coordinate system $\Phi$ and integrate the result over $D$. I'll demonstrate how to calculate the pullback of first summand $\frac{x dy \wedge dz}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}$ of $\omega$ to $D$ and leave the rest to you:

$$ \Phi^{*} \left(\frac{x dy \wedge dz}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} \right) = \frac{(\cos(\theta)\sin(\phi)) d(\sin(\theta)\sin(\phi)) \wedge d(\cos(\phi))}{1} = (\cos(\theta)\sin(\phi)) (\sin(\phi) d(\sin(\theta)) + \sin(\theta) d(\sin(\phi))) \wedge d(\cos(\phi)) \\ = (\cos(\theta)\sin(\phi))(\sin(\phi)\cos(\theta) d\theta + \sin(\theta)\cos(\phi) d\phi) \wedge (-\sin(\phi) d\phi) = \\ -\cos^2(\theta)\sin^3(\phi) d\theta \wedge d \phi.$$

The algorithm for computing the expression above is simply replacing $x,y,z$ with the appropriate expressions in terms of $\theta,\phi$ and then calculating the differentials using the exterior derivative calculus.