# Average volume of a spherical segment (frustum)

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.

• 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? Mar 16, 2023 at 17:31
• The right range should be $d\in[0,R-h]$. But even taking that into account the result doesn't match. Mar 16, 2023 at 17:34
• @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). Mar 16, 2023 at 21:46
• @Intelligentipauca Thanks for your correction! I’m going to edit immediately my original question. Mar 16, 2023 at 21:47

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.
• 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. Mar 17, 2023 at 6:54
• 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. Mar 17, 2023 at 18:35