I was reviewing the section on spherical coordinates in Griffith’s’ introduction to electrodynamics and I noticed that he includes unit vectors in the definition of an infinitesimal bit of area but not in that of an infinitesimal bit of volume.

The infinitesimal volume element $d\tau$, in spherical coordinates, is the product of the three infinitesimal displacements: $$d\tau=dl_r\,dl_\theta\,dl_\phi=r^2\sin\theta\,dr\,d\theta\,d\phi.\tag{1.69}$$ I cannot give you a general expression for surface elements $d\mathbf{a}$, since these depend on the orientation of the surface. You simply have to analyze the geometry for any given case (this goes for Cartesian and curvilinear coordinates alike). If you are integrating over the surface of a sphere, for instance, then $r$ is constant, whereas $\theta$ and $\phi$ change (Fig. 1.39), so $$d\mathbf{a}_1=dl_\theta\,dl_\phi\hat{\mathbf{r}}=r^2\sin\theta\,d\theta\,d\phi\,\hat{\mathbf{r}}.$$ On the other hand, if the surface lies in the $xy$ plane, say, so that $\theta$ is constant (to wit: $\pi/2$) while $r$ and $\phi$ vary, then$$d\mathrm{a}_2=dl_r\,dl_\phi\hat{\boldsymbol{\theta}}=r\,dr\,d\phi\,\hat{\boldsymbol{\theta}}.$$

Why is this? Neither volume nor area are vector quantities so I would think the infinitesimal of both shouldn’t need to have a unit vector. What would be undesirable about dropping the unit vector from $dA$?

  • 3
    $\begingroup$ if you really care, look into David Hestene's work on "Geometric Algebra"...it's too involved for an answer here, but it explains volumes elements, surface elements, and any other relevant elements in a manner that is independent of the number of dimensions. It really is the best way to understand these geometric objects, but it isn't generally taught. $\endgroup$
    – JEB
    Commented Jun 21 at 6:18

2 Answers 2


There is a sense in which area is a vector quantity, because a 2D surface embedded in a 3D space has not only some "size" but also an orientation. The area vector has a magnitude given by the scalar area of the surface, and a direction given by the orientation of the surface (normal to it).

Certainly in some situations you do not require information about the orientation of the surface. But in electrodynamics, that will frequently be critical information. You will need it to properly compute fluxes across surfaces, for example.

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    $\begingroup$ "area" is rather an anti-symmetric tensor than a vector. $\endgroup$
    – hyportnex
    Commented Jun 20 at 21:16
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    $\begingroup$ @hyportnex, if you want to be picky, it is a 2-form :-). All three (vector, tensor, 2-form) are vectors in an abstract sense $\endgroup$
    – Cryo
    Commented Jun 20 at 21:30
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    $\begingroup$ @cryo I am not trying to be picky just to let Noa know that "area" is a different beast from a directed segment, a vector, with which he would be familiar. $\endgroup$
    – hyportnex
    Commented Jun 20 at 21:40
  • $\begingroup$ @hyportnex Does drawing that distinction add value to the model? In classical electrodynamics, generally, no. So, traditionally, we don't do it. We aren't doing mathematics here, we're doing physics. $\endgroup$
    – John Doty
    Commented Jun 20 at 21:58
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    $\begingroup$ @JohnDoty in a vacuum it will not do anything for an antenna (B=H) but for a motor/dynamo/induction/Lorentz force, etc., it does help, methinks. This is an interesting subject but we are getting well beyond the scope of the original question. As I said above I only meant to warn, if that is the right word, Joa that a signed area naturally has a nonscalar nature and it is more complicated than a directed segment. Speaking from my own experience, I wish I had not been taught axial vectors before anti-symmetric tensors, and my comment was aimed to prevent that confusion I felt learning them. $\endgroup$
    – hyportnex
    Commented Jun 21 at 2:55

I think Riley's answer above is good already. All that I would add is that a good way to think about it is that you want to get back to a scalar when you are integrating. Why? Because often vector basis in physics are position-dependent. Spherical basis is a great example.

Lets say I have a vector field $\mathbf{A}=\mathbf{\hat{r}}$ and I want to integrate it over some area. What would this result even mean? Radial vector points in different directions in different places! We know how to integrate scalars, and the point of these vector-valued area elements, is to give you something to 'dot-product' with inside the integral to get to a scalar. So I could have an area element $\mathbf{\hat{x}}dxdy$, and now:

$$ \int_S \mathbf{A}.\mathbf{\hat{x}}dxdy $$

means something like the flux in the $\mathbf{\hat{x}}$ direction integrated over the area $S$.

More generally, this type of integration becomes much clearer when one uses differential forms, but they can be a bit of a heavy lift for an introductory EM course.

  • $\begingroup$ should "flux in the x direction along the area S" read "flux in the x direction through the area S" ? $\endgroup$
    – user45664
    Commented Jun 21 at 19:36
  • $\begingroup$ @user45664, I suppose if it is flux we are imagining some sort of a flow, so yes. Thanks $\endgroup$
    – Cryo
    Commented Jun 21 at 19:53

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