Is $d\sigma$ actually a differential? In what sense? When computing a surface integral, say
$ \displaystyle\int_S f(x) \, d\sigma $
we parameterize the surface $S$ by $r(u,v)$ and obtain $d\sigma = \vert r_u \times r_v \vert \, du\, dv$ to convert the integral into a double integral
$ \displaystyle\int_c^d\int_a^b f(r(u,v)) \, \vert r_u \times r_v \vert \, du\, dv $
Can $d\sigma$ actually be viewed as a differential in some sense? If so, what are we taking the differential of?
To compare, I realize that for a line integral, if the line is parameterized by $r(t)$, then the differential in the line integral satisfies $dr=r'\,dt$, which actually seems like taking a differential in the "traditional" sense, or even in the sense of differential forms.
But I guess I'm not sure exactly what $\sigma$ represents. Is it a form that one can take a differential of? I'm not sure how I would even compute $d\sigma$ for an integral in higher dimensions (say to compute the surface area of an $N$-dimensional sphere).
I also know that, to do such a computation, there is a formula
$\displaystyle\int_{S^{N-1}} d\sigma = N \displaystyle\int_{B^N} dx$
which seems related to generalized Stokes' theorem, but I'm not sure how to connect the dots, since I don't know how to take the differential of $d\sigma$. Is $d(d\sigma)=N\,dx$? That doesn't feel right...
 A: We can ask a similar question about line integrals. For example, when working with the unit circle $O$ we often use the differential form $\mathrm{d} \theta$. For example, the perimeter of the circle is
$$ 2 \pi = \oint_O \mathrm{d}\theta $$
There isn't a scalar field $\theta$ on the circle that $\mathrm{d} \theta$ would be the differential of (proof: if $f$ is a scalar field, then $\oint_O \mathrm{d}f = 0$).
So in that sense, the name $\mathrm{d} \theta$ for this 1-form is sort of an abuse of notation, or maybe it's better described as a different dialect of mathematical language; i.e. it should be read more as "an infinitesimal angle measure element" than as "the differential of angle".

However, locally on the circle, we can define a function $\theta$ such that $\mathrm{d} \theta$ is the differential of $\theta$; e.g. in the right half-plane we can define $\theta$ to be angular position ranging from $-\pi/2$ to $\pi/2$ in the usual way, and then $\mathrm{d}\theta$ is indeed the differential of $\theta$.
The same thing happens with surface integrals. The notation of differential can be applied to one-forms: the relevant identities are
$$ \mathrm{d}(f \, \mathrm{d}g) = \mathrm{d}f \mathrm{d}g 
\qquad \qquad \mathrm{d}f \mathrm{d}f = 0$$
In a similar fashion to the case of angle, any smooth 2-form on a smooth surface can be written as a differential; there does exist a differential 1-form $\sigma$ such that $\mathrm{d} \sigma$ is the differential of $\sigma$. But just like angle, you can't expect to be able to consistently define $\sigma$ on the entire surface simultaneously.
The reason why this should be true is that by a change of coordinates you can reduce the general case to the case of the plane, and you can always write $f(x,y) \, \mathrm{d}x \mathrm{d}y$ as a differential:
$$ f(x,y) \, \mathrm{d}x \mathrm{d}y
= \mathrm{d} \left( \left( \int f(x,y) \, \mathrm{d}x \right) \mathrm{d}y
\right)
$$
where the displayed integral is meant in the usual way as the antiderivative with respect to $x$ while holding $y$ fixed.
