[Math] Proof of coarea formula

analytic geometrymultivariable-calculus

I want to prove the coarea formula
\operatorname{Vol}(M) = \int_M d\operatorname{Vol}_M = \int_{-\infty}^\infty \frac{1}{|\nabla f|} \operatorname{area }(f^{-1}(t)) dt
where $f\colon M \rightarrow {\Bbb R}$ is a smooth function.

Here I have a question. The above formula says that $|\nabla f|$ is constant on $f^{-1}(t)$. So how can we prove this ? Thank you in advance.

Correction : I checked that the following equality is right (See 85 page in the book Eigenvalues in Riemannian geometry – Chavel)

$$\operatorname{Vol}(M) = \int_M d\operatorname{Vol}_M = \int_{-\infty}^\infty \int_{f^{-1} (t) } \frac{1}{|\nabla f|} d\operatorname{area}(f^{-1}(t)) dt$$

I believe that this formula is the generalization of arc-length parametrization. But (1) I cannot explain clearly and (2) I cannot prove the above inequality. So if you have an interests in this please help me in (1) and (2).

Best Answer

If $M$ is $n$-dimensional Riemannian manifold and if $\Omega$ is a domain in $M$ with a compact closure, then assume that for a smooth function $f: M\rightarrow {\bf R}$, $f^{-1}(t)$ is hypersurface in $\Omega$. Then for $F: M\rightarrow {\bf R}$, $$ \int_{ \Omega} F d{\rm Vol} =\int_{(-\infty,\infty)} \int_{S_t} \frac{F}{|\nabla f|} d{\rm Vol}_{S_t} dt $$ where $$S_t:= f^{-1}(t) \cap \Omega $$

Proof : Consider a vector field $\frac{\nabla f}{|\nabla\ f|^2} $ whose integral curve is $\psi (t,x)$. Then $\frac{\partial }{\partial t} f\circ \psi(t,x) =1$ so that $f\circ\psi (t,x) - f\circ \psi(x,0) =\int_0^t 1=t$ Hence $$ \psi (t,S_{t_0}) =S_{t_0+t} $$

Note that for small $t$, volume between $S_{t_0},\ S_{t_0+t}$ in $\Omega$ is approximately $$\int_{S_{t_0}} {\rm length}\ \psi|_{[t_0,t_0+t]} d{\rm Vol}_{S_{t_0}} \sim \int_{S_{t_0}} \frac{t}{|\nabla f|} d{\rm Vol}_{S_{t_0}} $$ By Cavalieri's principle, we complete the proof.

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