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 $dA$?
