# Volume element in spherical coordinates

In spherical coordinates, we have

$x = r \sin \theta \cos \phi$;

$y = r \sin \theta \sin \phi$; and

$z = r \cos \theta$; so that

$dx = \sin \theta \cos \phi\, dr + r \cos \phi \cos \theta \,d\theta – r \sin \theta \sin \phi \,d\phi$;

$dy = \sin \theta \sin \phi \,dr + r \sin \phi \cos \theta \,d\theta + r \sin \theta \cos \phi \,d\phi$; and

$dz = \cos \theta\, dr – r \sin \theta\, d\theta$

The above is obtained by applying the chain rule of partial differentiation.

But in a physics book I’m reading, the authors define a volume element $dv = dx\, dy\, dz$, which when converted to spherical coordinates, equals $r \,dr\, d\theta r \sin\theta \,d\phi$. How do the authors obtain this form?

• Have you read this? en.wikipedia.org/wiki/Volume_element – Siminore May 7 '14 at 9:13
• Oh, I hadn't read that. Thank you. I don't understand this part though: "...the volume element changes by the Jacobian of the coordinate change". I admit I haven't encountered the Jacobian before, but why would the volume element change by that? And if $dv = dxdydz$, what would I be doing wrong if I simply multiply out the $dx dy dz$ which were obtained in my post? – Train Heartnet May 7 '14 at 9:42
• Everything becomes clear if you study analysis on manifolds: you will learn the precise meaning of the volume element and how it behaves under changes of coordinates. It all boils down to the concept of differential form. I can suggest the little book by M. Do Carmo, Differential forms, or the book by Munkres, Analysis on manifolds. Finally, no, you can't just multiply $dx$, $dy$ and $z$. You must compute $dx \wedge dy \wedge dz$. – Siminore May 7 '14 at 11:27

$dV=dxdydz=|\frac{\partial(x,y,x)}{\partial(r,\theta,\phi)}|drd\theta d\phi$
• And $|\frac{\partial(x,y,x)}{\partial(r,\theta,\phi)}|= r^2 \sin\phi$..... – освящение Apr 8 '18 at 17:43