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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?

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  • $\begingroup$ @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
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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.

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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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