Let me motivate my question with this example.

The volume integral of a ball $\int_{B(0,R)} dx$ can be written as an integral over the surface of balls, i.e.

$$\int_{B(0,R)} dx = \int_0^R \int_{\partial B(0,r)}dS dr.$$

This shows, that the derivative w.r.t. $R$ is just the surface-integral

$$\frac{d}{dR} \int_{B(0,R)} dx = \int_{\partial B(0,R)}dS$$

Now, what happens if we generalize this: Let $F: \mathbb{R}^n \rightarrow \mathbb{R}$ and suppose that $\lambda( F^{-1}(-\infty, R]) < \infty$ for all $R$ then I see two ways to generalize my example:

(1) $\mu((-\infty,R]):=\int_{F^{-1}(-\infty, R]} dx$ defines a measure. The measure $\mu$ is a.c. with respect to the Lebesgue measure, i.e. $\mu(A)= \int_{A} fdx$ for some measurable $f$. In my example, the function $f(r)= \int_{\partial B(0,r)} dS.$

(2) But the most obvious generalization would be probably: $$\int_{F^{-1}(-\infty, R]} dx = \int_{-\infty}^{R} \int_{F^{-1}(\{r\})} \frac{1}{||\nabla F(x)||}dS(x) dr?$$

Apparently, (1) is more general than (2). I suspect (although do not know a proof of this) that (2) holds for submersions. Now my question is: How big is the difference between (1) and (2), i.e. does (1) only hold, if the representation in (2) holds almost everywhere? Under what precise conditions does (2) hold?

  • $\begingroup$ @NawafBou-Rabee yes, it is about a similar question, but this is completely unrigorous. Unfortunately, I could not find a reference in the math literature that studies this question (although it is a very classical one) , so I wanted to ask here about this. $\endgroup$ – Frederique Sep 23 '16 at 23:36
  • $\begingroup$ thanks for the reference, but I guess it does not really adress the question $\endgroup$ – Frederique Sep 23 '16 at 23:50
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    $\begingroup$ This looks like the coarea formula. $\endgroup$ – Neal Sep 24 '16 at 0:14

You need the coarea formula: $$ \int_E g(x)|\nabla u(x)|dx= \int_{-\infty}^{+\infty} \left( \int_{\{x\in E\ :\ u(x)=t\}} g(x)\ dH^{n-1}_ x \right) dt $$ where $E$ is an open subset of $R^n$, $u:E\to R$ is a Lipschitz function, $g\in L^1$ and $dH^{n-1}$ is the Hausdorff measure (surface measure). If you pick $g(x)= |\nabla u(x)|^{-1}$ and $E=\{u(x)<R\}$ you get your formula. The assumptions on $g,u,E$ can be further weakened. It should be pretty straightforward to collect information on the coarea formula on the web.


This is Federer's coarea formula. You find it in this classical but dense book:

Federer, Herbert (1969), Geometric measure theory, Die Grundlehren der mathematischen Wissenschaften, Band 153, New York: Springer-Verlag New York Inc., pp. xiv+676.

This question on derivatives of volumes prompts me with the following curiosity. Let $S_n$ be the surface area of the unit hypersphere in $\mathbb{R}^n$ (avoiding the topologists notation $\mathbb{S}^{n-1}$, for convenience), and treat $S_n$ as a function of a continuous variable $n$ (relying on Euler's $\Gamma$-function). Now, setting the derivative $\frac{dS_n}{dn}=0$ leads to $n=7.35\dots$; hence the "amusing" fact that the $7$-dimensional unit hypersphere has the maximum surface area.

On the other hand, we have some striking results such as the first among the spheres having exotic structures is the $7$-dim sphere (Milnor). Plus, while no even dimensional spheres are parallelizable, there are only $1$-d, $3$-d and $7$-d among the odds. Then it stops.

My question is: what is so special about the $7$-dimensional sphere? I wonder.


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