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Suppose $\omega$ is a bounded shape in $\Bbb{R}^3$ and that $\{z : (x,y,z) \in \omega \}=[0,T]$ (that is, the shape is exactly contained in the band $\{z \in [0,T]\}$. If we denote by $\omega_t = \{(x,y,t) \in \omega\}$ the slice of $\omega$ at height $t \in [0,T]$ and $A_t$ the area of $\omega_t$ it is well known that $\int_0^T A_t dt$ gives the volume of $\omega$.

Is there a way to compute $\displaystyle \int_0^T (A_t)^2 dt$ explicitly?

If such a formula exists, it would be useful in an application where I need that the areas of the slices along a certain direction have the smallest variation.

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    $\begingroup$ It’s not clear to me what sort of formula you’re looking for — for general $\omega$, what kind of formula could one hope for that would be more explicit than $\int_0^T (A_t)^2 dt$? Can you give an example to illustrate what you’re looking for — i.e. a formula to “explicitly compute” some other integral associated to an object, in the sense you have in mind? $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented May 4, 2018 at 13:43
  • $\begingroup$ @PeterLeFanuLumsdaine: In the case where the integrand is $A_t$, the integral is equal to the volume. For me, a satisfactory answer would be that the integral can be computed by integrating a function on the whole body $\omega$, rather than integrating with respect to the height parameter. I'm not sure such an answer exists. At least I didn't find one, but before deciding that this isn't possible, I wanted to ask the question. $\endgroup$
    – Beni Bogosel
    Commented May 4, 2018 at 15:14
  • $\begingroup$ @BeniBogosei: Saying that “in the case where the integrand is $A_t$, the integral is equal to the volume” seems more of a geometric interpretation than a way to explicitly compute the integral? If you want an integral over the whole body recovering your integral-of-squares-of-slices, then you can take $\int_{(x,y,z) \in \omega} A_z dx dy dz$; this again doesn’t seem to me like it helps to “explicitly compute” the original, but I still don’t follow what kind of formula you’re hoping for, so maybe it is of use? $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented May 4, 2018 at 17:36
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    $\begingroup$ It is clear from the case of a rectangle that this integral is not rotation invariant, so it doesn't compute a geometric quantity. $\endgroup$
    – Ben McKay
    Commented May 5, 2018 at 6:16

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