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I have always pictured volume element as a small cuboid in with volume $dx dy dz$. however in curvilinear system, how would the shape of this volume element be?

I mean in spherical polar coordinate system, how the shape of this volume element be visualized (is it a small sphere whose integration give the volume of object or same), or my idea of 3D Cartesian coordinate absolutely wrong.

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    $\begingroup$ It's still a rectangular prism, but rotated depending on its location. goo.gl/AdoDu $\endgroup$
    – Mark Eichenlaub
    Dec 22, 2010 at 9:20
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    $\begingroup$ In general system it need not be rectangular. If $e_1, e_2, e_3$ are tangential vectors to the coordinates at some point then the volume element will be $V = \{x \in {\mathbb R}^3 \,|\, x = \sum_i t_i e_i,\, t_i \in [0, 1], \, i = 1,2,3 \}$ $\endgroup$
    – Marek
    Dec 22, 2010 at 10:10
  • $\begingroup$ More a math question actually. Not sure wether to close since it is an easy one and it is also relevant to physics. $\endgroup$
    – Raskolnikov
    Dec 22, 2010 at 13:29

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the easiest way of picturing the volume element is to start in a point $x\in M$ given in local coordinates $x^\mu$ and go along a (maybe curved) line $x^\mu + dx^\mu$.

So, lets do this! In a two dimensional polar coordinate system you can start at some point that is given by a radius $r = r_0$ and some angle $\varphi = \varphi_0$. Now, you draw a small arc segment from $(r_0, \varphi_0)$ to $(r_0, \varphi_0 + d\varphi)$ where $d\varphi$ is just a small angle and you hold the radius constant.
Then, you do the same from $(r_0, \varphi_0)$ to $(r_0 + dr, \varphi_0)$. Now you mark the point $(r_0 + dr, \varphi_0 + d\varphi)$ and connect this point (in the same manner as you drew the segments before) to the endpoints of your lines.

Now, can you calculate the surface area of this little element?

If you have done this, can you relate your calculation to the volume form $\mathrm{vol} = \sqrt{g} dy^1\wedge\dots\wedge dy^n = r dr \wedge d\varphi$?

Sincerely

Robert

PS.: I left out the orientation of the manifold for clearity.

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    $\begingroup$ I would add that in general one doesn't need a metric; volume form is a structure that can be defined independently of it. Nevertheless, it's often the case that volume form is canonically attached to some other structure on the manifold. One prominent example of this besides metric volume form is a phase volume form on $n$-dim symplectic manifolds defined as $\omega^{\wedge n}$ where $\omega$ is the symplectic form. $\endgroup$
    – Marek
    Dec 22, 2010 at 12:09
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    $\begingroup$ Correction: $2n$-dim manifold. $\endgroup$
    – Marek
    Dec 22, 2010 at 12:15
  • $\begingroup$ @Marek: Thank you for the additional information. I think we are pretty safe here restricting the discussion to the meaning of volume for the case of some Riemannian manifold $(M,g)$ :) $\endgroup$
    – Robert Filter
    Dec 22, 2010 at 14:01
  • $\begingroup$ Nice answer @robert. However, a person unfamiliar with the $\wedge$ product might find the last part confusing. $\endgroup$
    – user346
    Dec 22, 2010 at 21:49
  • $\begingroup$ @Robert: sure. One can probably even safely assume Euclidean space. Although once one starts working in general coordinates it's pretty natural to generalize to arbitrary manifolds as you did. By the way, I agree with space_cadet, nice answer :-) $\endgroup$
    – Marek
    Dec 22, 2010 at 22:06

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