Questions on calculating volume using n-1 forms Is there an n-1 form on $R^n$ which calculates the volume of n-manifolds? Similarly, is there such a 1 form on $S^2$, and $RP^2$? I thought this has something to do with the orientation, is that right? 
 A: The question is a bit vague so I will try to make the best of it.  Suppose that $(M, g)$ is a connected, noncompact  oriented, $n$-dimensional  Riemann manifold and $dV_g\in\Omega^n(M)$ is the associated  volume form.    Suppose  that $D\subset M$ is an open, precompact subset   of $M$ with smooth boundary $\partial D$.  Since $M$ is noncompact we have  $H^n(M,\mathbb{R})=0$ and we deduce  that there exists  an $(n-1)$ form $\eta$ such that $d\eta=dV_g$.  Stokes' theorem then implies  that
$$ \mathrm{vol}(D, g)=\int_D dV_g = \int_{\partial D} \eta. $$
For example, if $(M,g)$ is the Euclidean space $\mathbb{R}^n$, then we can take
$$\eta=\frac{1}{n}\sum_{k=1}^n (-1)^{k-1} x^k dx^1\wedge \cdots \wedge\widehat{dx^k}\wedge \cdots dx^n. $$
If $M$ is compact, and $x\in M\setminus \bar{D}$, then $M\setminus x$ is noncompact and arguing as above  we can find $\eta_x\in \Omega^{n-1}(M\setminus x)$ such that $dV_g|_{M\setminus x}=d\eta_x$ and 
$$ \mathrm{vol}(D, g)= \int_{\partial D} \eta_x. $$
A: No. $n-1$-forms are equivalent to vector fields, and the measurement of "volume" you get is the integral of the normal vector dot the vector field, not the integral of the length of the normal vector, which measures volume.
The easiest way to see this is in $\mathbb R^2$. You want to measure the length of a curve, say parameterized by a coordinate $t$. The length is the integral of $\sqrt{(dx/dt)^2+(dy/dt)^2dt}dt$. The integral by the one-form $adx+bdy$ would be $a(dy/dt)-b(dx/dt)$, which looks very different.
