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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?

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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

$$ \mathrm{vol}(D, g)= \int_{\partial D} \eta_x. $$

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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.

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Do you mean that $n-1$-forms are equivalent to vector fields? $n$-forms are equivalent to functions, aren't they? –  MTS Apr 23 '12 at 22:31
    
On $\mathbb R^n$, you have a canonical $n$-form, the volume form, which gives a natural isomorphism between between $n$-forms and functions (and vector fields and $n-1$-forms, and so on). –  Will Sawin Apr 24 '12 at 2:53

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