## The volume of the intersection of balls

Fix two positive numbers ${R}$ and ${r}$. The volume of the intersection of balls of radii ${R}$ and ${r}$ is a function of the distance ${d}$ between their centers. Is it a decreasing function?

Wolfram MathWorld gives a formula for this volume in the three-dimensional case

$\displaystyle V(d) = \frac{\pi}{12 d} (R+r-d)^2(d^2+2rd-3r^2+2Rd+6Rr-3R^2)$

(Here and in the sequel it is assumed that ${d\le R+r}$.) From this formula it is not at all obvious that ${V(d)}$ is decreasing. And this is just for three dimensions.

Let’s begin with the one-dimensional case, then. Consider the intervals ${[-R,R]}$ and ${[d-r,d+r]}$ where ${d\le R+r}$. A point ${x}$ belongs to their intersection if and only if

$\displaystyle \max(-R,d-r)\le x \le \min(R,d+R)$

Hence, the length of the intersection is ${L(d)=\min(R,d+r)-\max(-R,d-r)}$. Even this simple formula does not quite make it evident that the function is decreasing for ${d>0}$. If ${d+r>R}$ this is clear enough, but the case ${d+r -d-r>-r}$.

Fortunately, Fubini’s theorem reduces the general case to the one-dimensional one. Take any line ${L}$ parallel to the line connecting the centers. The intersection of each ball with ${L}$ is a line segment whose length is independent of ${d}$. The distance between the midpoints of these segments is ${d}$; thus, the length of the intersection is a decreasing function of ${d}$.

Gromov’s note Monotonicity of the volume of intersection of balls (1987) deals with the more general case of ${k\le n+1}$ balls in ${\mathbb R^n}$: if two sets of balls ${B(x_i,r_i)}$ an ${B(x_i',r_i)}$ satisfy ${\|x_i'-x_j'\|\le \|x_i-x_j\|}$ for all ${i,j}$, then the volume of ${\bigcap B(x_i,r_i)}$ does not exceed the volume of ${\bigcap B(x_i',r_i)}$.

The paper ends with

Question: Who is the author of [this result]? My guess is this was known to Archimedes. Undoubtedly the theorem can be located […] somewhere in the 17th century.

Meanwhile, the corresponding problem for the volume of the union remains open.

## Rectangular boxes: handle with care

A rectangular box (aka parallelepiped) looks like a sturdy object:

But this particular box, with dimensions 7.147 by 6.021 by 4.095, took me the better part of an hour to figure out.

It was a part of a numerical methods assignment: find the dimensions of a box with given volume ${V}$, surface area ${S}$, and diameter ${D}$ (i.e., space diagonal). Algebraic approach leads to pretty ugly expressions, which is the motivation for a numerical method. Specifically, the task was to apply the Newton-Raphson method to the map

$\displaystyle F(x,y,z) = \begin{pmatrix} xyz-V \\ 2(xy+yz+xz)-S \\ x^2+y^2+z^2-D^2 \end{pmatrix}$

Of course, I understood that not every triple ${(V,S,D)}$ is attainable. Also realized that the Jacobian of ${F}$ is degenerate when two of the coordinates coincide, which is a problem for the method. So I thought: let’s generate some random ${x,y,z}$ values that are not too close to one another, and give students the resulting parameters ${V,S,D}$.

With ${x=7.147}$, ${y=6.021}$, and ${z=4.095}$ the parameters are ${V=176.216}$, ${S=193.91}$, and ${D=10.203}$. Sure, a little rounding can’t hurt when numbers are of this size and we are quite far from the critical points of ${F}$. So I put ${V=176}$, ${S=194}$ and ${D=10}$ in one of the versions of the assignment.

But the Newton-Raphson method would not converge… because no such box exists! The rounding did hurt after all.

This motivated me to describe all attainable triples ${(V,S,D)}$ explicitly, which ended up being less of a chore than I expected. It helps to realize that ${(x+y+z)^2 = D^2+S}$, which reduces the search to the intersection of the sphere ${x^2+y^2+z^2=D^2}$ with the plane ${x+y+z=\sqrt{D^2+S}}$. This is a circle (called ${C}$ below), and the allowed range for ${V}$ is between the minimum and maximum of ${xyz}$ on ${C}$.

This goes into Calculus 3 territory. Using Lagrange multipliers with two constraints looks like a tedious job. Instead, I decided to parametrize ${C}$. Its center is ${(c,c,c)}$ where ${c=\dfrac13\sqrt{D^2+S}}$. The radius is ${\displaystyle r = \sqrt{ D^2 - \frac13(D^2+S)}=\frac13 \sqrt{6D^2-3S}}$. We also need an orthonormal basis of the subspace ${x+y+z=0}$: the vectors

$\displaystyle \frac{1}{\sqrt{6}} \langle 2, -1, -1\rangle \quad \text{and}\quad \frac{1}{\sqrt{2}} \langle 0, 1, -1\rangle$

do the job.

So, the circle ${C}$ is parametrized by

$\displaystyle x = c+\frac{2r}{\sqrt{6}} \cos t \\ y = c-\frac{r}{\sqrt{6}} \cos t +\frac{r}{\sqrt{2}} \sin t \\ z = c-\frac{r}{\sqrt{6}} \cos t -\frac{r}{\sqrt{2}} \sin t$

This is not as bad as it looks: the product ${xyz}$ simplifies to

$\displaystyle xyz = c^3 - \frac{cr^2}{2} + \frac{r^3\sqrt{6}}{18} \cos 3t$

which tells us right away that the volume ${V}$ lies within

$\displaystyle c^3 - \frac{cr^2}{2} \pm \frac{r^3\sqrt{6}}{18}$

In terms of the original data ${S,D}$ the bounds for ${V}$ take the form

$\displaystyle \frac{5S-4D^2}{54}\sqrt{S+D^2} \pm \frac{\sqrt{2}}{54} (2D^2-S)^{3/2}$

(And of course, ${V}$ cannot be negative even if the lower bound is.) It is easy to see that ${2D^2-S\ge 0}$ with equality only for a cube; however ${2D^2-S}$ can be relatively small even when the box does not look very cube-ish. For the box pictured above, the tolerance ${\frac{\sqrt{2}}{54} (2D^2-S)^{3/2}}$ is approximately ${1.4}$; after rounding ${S\approx 194 }$ and ${D\approx 10}$ this drops to ${0.38}$, and the desired volume of ${176}$ is way out of the allowable range ${181\pm 0.38}$.

Yes, the set of attainable triples ${(V,S,D)}$ is quite thin. Parallelepipeds are fragile objects: handle them with care.

## Interpolating between a cone and a cylinder

This post is inspired by James Tanton’s video which is summarized in bulleted items below. For convenience normalize all solids to height ${1}$.

• Parallel cross-sections of a cylinder have constant area. Therefore its volume is ${\int_0^1 A z^0\,dz =Ah}$, where ${A}$ is the area of the base.
• Parallel cross-sections of a cone are scaled in both directions by the ${z}$-coordinate. This multiplies the sectional area by ${z^2}$ and therefore the volume is ${\int_0^1 A z^2\,dz =\frac{1}{3}Ah}$.
• There is something in between: we can scale cross-sections in one direction only. Then the sectional area is multiplied by ${z}$ and the volume is ${\int_0^1 A z\,dz =\frac{1}{2}Ah}$.

Taking the base to be a circle, we get this shape:

It looks… odd. Not even convex:

But cones and cylinders are convex whenever the base is a convex region. Can we interpolate between them within the class of convex sets?

The answer is: yes, if we use the Minkowski sum. Given two convex bodies ${A_0}$ and ${A_1}$ in ${{\mathbb R}^3}$, for ${0< t<1}$ define

$\displaystyle A_t =(1-t)A_0+tA_1 = \{(1-t)a_0+ta_1 : a_0\in A_0, a_1\in A_1\}$

Each ${A_t}$ is convex. If the sets ${A_0}$ and ${A_1}$ are cone and cylinder of equal height and with the same convex base ${B}$, then all ${A_t}$ have the same base and the same height.

Finding the volume of a convex body in Calculus II/III can be a tiresome exercise with trigonometric substitutions and whatnot. It may come as a surprise that the volume of ${A_t}$ is always a cubic polynomial in ${t}$, even if the convex sets ${A_0, A_1}$ are defined by some transcendental functions. I still do not find this fact intuitive.

Let’s check it on the example of ${A_0}$ and ${A_1}$ being a cone and a cylinder of height ${1}$ and radius ${r}$. Obviously, every ${A_t}$ is a solid of revolution. Its profile is the Minkowski sum of appropriately scaled triangle and rectangle. This is how I imagine it:

• Mark a point on the rectangle, for example the midpoint of the bottom edge.
• Move the rectangle around so that the marked point stays within the triangle.
• The region swept is the Minkowski sum.

The volume can now be computed directly. I used the venerable method of cylindrical shells to find the volume of ${A_1\setminus A_t}$:

$\displaystyle 2\pi \int_{tr}^t x(x/r-t)\,dx = \frac{\pi r^2}{3}(2-3t+t^3)$

hence

$\displaystyle \mathrm{Vol}\, (A_t) = \frac{\pi r^2}{3}(1+3t-t^3)$

It so happens that ${\mathrm{Vol}\,(A_t)}$ is a concave function in this example, although in general only its cubic root is guaranteed to be concave, by the Brunn-Minkowski inequality.

When ${t\approx 0.168}$, the volume of ${A_t}$ is equal to ${\pi r^2/2}$, same as the volume of James Tanton’s solid. Since ${t}$ is pretty small, the shape looks much like a cone.

When ${t\approx 0.347}$, the volume of ${A_t}$ is equal to ${2\pi r^2/3}$, which is exactly halfway between the volumes of cone and cylinder. The solid exhibits harmonious balance of both shapes.

## Archimedean volume ratio

And although he made many excellent discoveries, he is said to have asked his kinsmen and friends to place over the grave where he should be buried a cylinder enclosing a sphere, with an inscription giving the proportion by which the containing solid exceeds the contained.

(Quote from Plutarch; see references on the Archimedes site by Chris Rorres. Image from Wikipedia)

The proportion is $3:2$, as Archimedes discovered. More generally, let $A_n$ be the ratio of the volumes of $n$-dimensional unit ball $B^n$ and of the circumscribed cylinder $B^{n-1}\times [-1,1]$. This sequence begins with $A_1=1$, $A_2=\pi/4$, $A_3=2/3$, $A_4=3\pi/16$, $A_5=8/15$, $A_6=5\pi/12\approx 0.49\dots$. Note that beginning with $6$ dimensions, the ball occupies less than half of the cylinder. It is not hard to see that the sequence monotonically decreases to zero. Indeed, the ratio of $(n-1)$-dimensional cross-sections by the hyperplane $x_{n}=t$, for $-1\le t\le 1$, is $(1-t^2)^{(n-1)/2}$. Hence $A_n = \frac{1}{2} I_{(n-1)/2}$ where $I_p:=\int_{-1}^1 (1-t^2)^p\,dt$. The integrand monotonically converges to $0$ a.e. as $p\to\infty$, and therefore $I_p$ does the same.

The Archimedean ratio $A_n$ is rational when $n$ is odd, and is a rational multiple of $\pi$ when $n$ is even. Indeed, integrating $I_p$ by parts, we get
$\displaystyle I_p = \int_{-1}^1 (1-t^2)^p\,dt = 2p \int_{-1}^1 t^2(1-t^2)^{p-1}\,dt = 2p(I_{p-1}-I_{p})$
hence $I_p = \frac{2p}{2p+1}I_{p-1}$. In terms of $A_n = \frac{1}{2} I_{(n-1)/2}$, we have $A_{n}= \frac{n-1}{n}A_{n-2}$. Since $A_1=1$ and $A_2=\pi/4$ are known (the former being trivial and the latter being one of definitions of $\pi$), the claim follows.

Combining the monotonicity with the recurrence relation, we find that $\displaystyle \frac{n-1}{n} =\frac{A_{n}}{A_{n-2}} \le \frac{A_{n}}{A_{n-1}}\le 1$. Hence $\displaystyle \frac{A_{n}}{A_{n-1}} \to 1$, which yields the Wallis product formula for $\pi$. Indeed, the recurrence shows that for $n$ even, $\displaystyle \frac{A_{n}}{A_{n-1}}$ is a particular rational multiple of $\displaystyle \frac{A_{2}}{A_{1}} = \frac{\pi}{4}$.

The above is essentially the inductive proof of Wallis’s formula I remember from my calculus (/real analysis) class. Unfortunately, between integration by parts and the squeeze lemma we lose geometry. In geometric terms, Wallis’s formula says that the volumes of $B^{n+1}\times B^{n-1}$ are $B^n\times B^n$ are asymptotically equivalent: their ratio tends to $1$. Is there a way to see that these solids have a large overlap? Or, to begin with, that $B^{n+1}\times B^{n-1}$ has the smaller volume? I don’t see this even with $B^3\times B^1$ vs $B^2\times B^2$. By the way, the latter is the bidisc from complex analysis.

Aside: Wallis’s formula for statistics students by Byron Schmuland presents a nice 2-page proof (which uses the relation between $\pi$ and the Gaussian).

Added: Another way to generalize the Archimedean volume ratio to higher dimensions is to divide the volume of $B^{2n+1}$ by the volume of $B^{n}\times B^{n+1}$. The quotient is always rational, namely $\displaystyle \frac{(n+1)!}{(2n+1)!!}$. The sequence begins with $\displaystyle 1, \frac23, \frac25, \frac{8}{35}, \frac{8}{63}, \frac{16}{231}, \frac{16}{429}$ and obviously tends to zero. It does not appear to be related to $\pi$.