Average volume of a spherical segment (frustum)

Question

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.

Answer 1

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.

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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.