Is there an equivalent to the line integral for area in $\mathbb R^3$? So the following line integral is known to give the area of some figure in $\mathbb R^2$:
$$A=\frac{1}{2}\oint_C x\;dy-y\;dx$$
Is there an equivalent expression in $\mathbb R^3$ for volume?
 A: You can use the Divergence Formula for $\vec{F} = \langle x,y,z \rangle$ and you will have:
$$\frac 13 \iint _A \vec{F}\cdot \vec{n} \, dA = \frac 13 \iiint _V \operatorname{div} \vec{F} \, dV = \iiint_V dV = V$$
In a similar manner you can use the same trick for $\vec{G} = \langle x,0,0 \rangle$
$$\iint _A \vec{G}\cdot \vec{n} \, dA =  \iiint _V \operatorname{div} \vec{G} \, dV = \iint_V dV = V$$
But keep in mind the formulae aren't really that simple as in $\mathbb{R}^2$, as after all it depends on the normal vector. So sometimes we have to make the right choice for the vector field in order to get an easier computation problem.
A: First, let’s look at how it is that the (signed) area of the region $R$ with boundary $C$ can be computed via $\frac12\int_Cx\,dy-y\,dx$. Stokes’ theorem tells us that this integral is equal to $$\frac12\int_Rd(x\,dy-y\,dx)=\frac12\int_{\partial R}2\,dx\wedge dy=\int_Cdx\wedge dy.$$ If we apply Stokes’ Theorem in the other direction, it tells us that we can use any antiderivative of $dx\wedge dy$. E.g., $x\,dy$ works just as well as an integrand and can often be more convenient for computation. In general, we can use any convenient 1-form $\omega=x\,dy+\tau$, where $d\tau=0$ ($\tau$ is analogous to the constant of integration in single-variable indefinite integrals).  
This has an obvious generalization to $n$ dimensions: the integrand $\omega$ must be such that $d\omega=dx^1\wedge\cdots\wedge dx^n$. One possible such $(n-1)$-form is the generalization of the form used to calculate areas: $$\frac1n(x^1\,dx^2\wedge\cdots\wedge dx^n+x^2\,dx^3\wedge\cdots\wedge dx^n\wedge dx^1+\cdots+x^n\,dx^1\wedge dx^2\wedge\cdots\wedge dx^{n-1})\tag{*}$$ which you can find via an educated guess or the algorithm described here. In three dimensions, this is $$\int_Vdx\wedge dy\wedge dz=\frac13\int_{\partial V}x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy.$$ This is just an instance of the divergence theorem (a.k.a. Gauss’ theorem) expressed in terms of differential forms.  
On the other hand, just as in the two-dimensional case, there are other integrands that can be much more convenient for computation. For instance, $x^1\,dx^2\wedge\cdots\wedge dx^n$ is also an antiderivative of the volume element, as are all of the other terms inside of the parentheses in (*). Indeed, you can think of that entire expression as the average of those individual antiderivatives, each one in the direction of a particular basis vector. There are other linear combinations of these terms that might also be more convenient in various situations.
The integrand $\frac12(x\,dy-y\,dx)$ has a nice geometric interpretation as the area of the triangle with vertices $(0,0)$, $(x,y)$ and $(x+dx,y+dy)$, and the $(n-1)$-form generalization of it can also be interpreted geometrically. After parameterization, the terms in the sum can be computed as determinants, which in turn have an interpretation as volumes of certain paralleletopes that cover the volume being computed.
