# Triple Integral: Coordinate change from polar to spherical $(dx\wedge dy\wedge dz)\rightarrow(d\rho\wedge d\phi\wedge d\theta)$

Let $$x=\rho \sin(\phi)\cos(\theta), y=\rho \sin(\phi)\sin(\theta), z=\rho \cos(\phi)$$. Then $$dx=d(\rho \sin(\phi)\cos(\theta))=\sin(\phi)\cos(\theta)d\rho-\rho \sin(\phi)\sin(\theta)d\theta+\rho \cos(\phi)\cos(\theta)d\phi$$ $$dy=d(\rho \sin(\phi)\sin(\theta))=\sin(\phi)\sin(\theta)d\rho+\rho \sin(\phi)\cos(\theta)d\theta+\rho \cos(\phi)\sin(\theta)d\phi$$ $$dz=d(\rho \cos(\phi))=\cos(\phi)d\rho-\rho \sin(\phi)d\phi$$ I then wedged $$dx\wedge dy$$ first and got: $$dx\wedge dy=\rho \sin^2(\phi)d\rho d\theta-\rho^2\sin(\phi)\cos(\phi)d\theta d\phi$$ So then $$dx\wedge dy\wedge dz$$ should be: $$dx\wedge dy\wedge dz=(-\rho^2\sin^3(\phi)-\rho^2\sin(\phi)\cos^2(\phi))d\rho d\theta d\phi=-\rho^2\sin(\phi)d\rho d\theta d\phi$$ But I know it's supposed to be positive. I am sure I made some minor mistakebut I've been staring at this for sometime and am starting to wonder if I'm conceptually mis\sing something bigger that is going on.

It just dawned on me... do i simply have a slightly switched order of the differentials than the formula that is normally given with the positive value?

• I suspect so. $d\rho \wedge d\theta \wedge d\phi = -d\rho \wedge d\phi \wedge d\theta$, and the latter should be the correct ordering considering physical intuition that $\hat{\rho}$, $\hat{\phi}$, $\hat{\theta}$ should form a right-handed orthonormal basis in this order. – Sangchul Lee Apr 27 at 1:48
• so the standard notation for spherical coordinates is $d\rho\wedge d\phi\wedge d\theta$ for spherical coordinates? – joseph Apr 27 at 2:45