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SE users, I was trying to evaluate the average volume of a spherical segment (frustum), as that shown in this Wolfram link. I understood that in general the volume of a frustum of a sphere with radius $R$, given its height $h$ and the distance from the center to the start of the segment $d$, is: $$ V=\pi h\left(R^2-d^2-hd-\frac{1}{3}h^2\right). $$ Now, I need to calculate the average of this volume $V$ over the variable $d$, but I have difficulties on understand how to proceed. The expected result should be: $$ \left\langle V\right\rangle=\frac{4}{3}\pi \frac{R^3h}{2R+h}, $$ as stated in the Supplementary Information (Equation $1$) of this Nature article. I tried to work as if the average was intended as an ensemble average (in a probabilistic sense), considering $d$ to be a random real variable ranging in the interval $\left[0,R-h\right]\subset \mathbb{R}$. In particular, I supposed that $d$ satisfied a continuous uniform distribution because all points in the finite interval are equally likely. In this case, the mean (first raw moment) of the continuous uniform distribution is $$ E\left[d\right]=\frac{R-h+0}{2}=\frac{R-h}{2}, $$ but this doesn't give the expected result so I concluded that was a wrong reasoning. Later, I tried to interpret the problem in a geometrical way, by intending the average of $V$ over $d$ as an arithmetic mean (i.e. a continuous integral) over the possible values of $d$: $$ \left\langle V\right\rangle=\int_{0}^{R-h}V(d) \ \mathrm{d}(d), $$ but again I didn't find the expected result (the value of the integral was even negative). At this point, I think I must invoke a cylindrical (or spherical) representation of the variables in exam in order to appropriately rewrite the above integral, but I don't know if this is the right way and even how to effectively do it.

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    $\begingroup$ In the supp. notes, the author says $\langle\cdot\rangle$ denotes "the average of a probability", whereas it seems you've interpreted it as plain old average/expectation. Do you know if "average probability" is defined anywhere else in the article, or if the author is using it interchangeably? $\endgroup$
    – user170231
    Mar 16, 2023 at 17:31
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    $\begingroup$ The right range should be $d\in[0,R-h]$. But even taking that into account the result doesn't match. $\endgroup$ Mar 16, 2023 at 17:34
  • $\begingroup$ @user170231 Unfortunately, there is no place either in the article or in the supp. notes in which is specified what “average” means. I supposed that the probability in question was that of the variable $d$, i.e. a continous uniform probability distribution. That seemed to be the easiest way to reason in statistical terms but starting from the geometry of the problem (something like models of static polymers). $\endgroup$ Mar 16, 2023 at 21:46
  • $\begingroup$ @Intelligentipauca Thanks for your correction! I’m going to edit immediately my original question. $\endgroup$ Mar 16, 2023 at 21:47

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Your formula for the frustum volume $$V=\pi h\left(R^2-d^2-hd-\frac{1}{3}h^2\right)\tag1$$ is correct but the parameter $d$, the distance from the equatorial plane to the "lower" surface of the segment, must be a signed value; otherwise you won't be able to handle the case where the frustum contains the equator. You can check that formula (1) is valid for $d\in[-R, R-h]$. To get the average volume you would integrate (1) against an appropriate density function $f(d)$ for the distribution of $d$.

That being said, unless I'm misunderstanding something, I cannot see a density $f$ that would give an average volume of $$\left\langle V\right\rangle=\frac{4}{3}\pi \frac{R^3h}{2R+h}\tag2$$ as claimed in the article. The formula (2) appears to be incorrect, for it gives the wrong value for some obvious special cases. Consider $h=2R$: If the segment consumes the entire sphere, its volume should equal the sphere's volume, whereas (2) gives half that. And when $h=R$, every segment of thickness $h$ consumes at least half the volume of the sphere; but (2) asserts that the expected volume of the segment is one-third of the volume of the sphere.


ADDED: If we assume $f(d)$ has uniform distribution over $d\in[-R, R-h]$, then integrating (1) against this choice for $f$ and dropping higher order terms in $h$ yields $$\langle V\rangle \sim \frac 43\pi\frac{R^3h}{2R -h}$$ as $h\to0$. So there may be a typographical error in the formula stated in the article.

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  • $\begingroup$ Thank you for answering my question! With your suggestions I’ll try to understand what kind of erroer lies under your formula (2). By reasoning backwards, I hope to understand what probability density function to use for the random variable $d$ that now finally ranges in the right interval. $\endgroup$ Mar 17, 2023 at 6:54
  • $\begingroup$ Thanks to your input I tried to manually compute the integral of the function $\frac{{\pi}h\cdot\left(-d^2-hd-\frac{h^2}{3}+R^2\right)}{R-h-(-R)}$ in the interval $[-R,R-h]$, considering that the continuous uniform probability distribution function $f(d)=\frac{1}{R-h-(-R)}$. The computation gives $\frac{{\pi}h\cdot\left(h^3-4Rh^2+8R^3\right)}{6\left(2R-h\right)}$ and as you said, neglecting the first two terms in the bracket (of order higher than $1$ in $h$), we recover the exact result of the article with a change in its sign. $\endgroup$ Mar 17, 2023 at 18:35

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