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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).

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    If you just write out words like Vol and area, they get interpreted as juxtapositions of variable names, and formatted accordingly (e.g. italicized). To get the right font and spacing, you need to use constructions like `\operatorname{Vol}`.2012-12-12
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    I think that the claim is false if $f^{-1}(t)$ is not connected. For example, if $$f(x, y)=\left(x^2+y^2\right)\left[\left(x-1\right)^2+y^2\right],$$ then $(1/2, 0)$ and $(1/2+ \sqrt{2}/2, 0)$ are on the same level set $f^{-1}(1/16)$ but $$\nabla f(1/2, 0)=0,\quad \nabla f(1/2+\sqrt{2}/2, 0)\ne 0.$$ Unless I have made a computational error, of course.2012-12-12
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    You are right. $\nabla f(1/2,0) = (1/4,0)$ $\nabla f(1/2 + \sqrt{2}/2,0) = (\sqrt{2}/2,0)$. Another example is $f(x,y)=x^2/a^2 + y^2/b^2$ $(a\neq b)$2012-12-13
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    @joriki : Thank you. I editted.2012-12-13
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    You edited one out of three occurrences of the problem.2012-12-13
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    I understand exactly what you said. Thank you. I fixed them.2012-12-13

1 Answers 1

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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.