Video 1, chapter 4

The Napkin Ring Paradox

Why napkin rings of equal height have equal volume whatever the size of the ball, explained through the napkin ring problem, cross-sections, the area of an annulus, Cavalieri's principle and the volume of a sphere.

  • Video 1, chapter 4
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Video 1, chapter 4

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Drill a straight hole through the center of a ball the size of a grapefruit, and another through a ball the size of a beach ball, with the holes chosen so that both leftover rings are 6 centimeters tall. The ring from the big ball reaches much farther around, so it seems to contain more material. It does not, because both rings have the same volume, 36π36\pi cubic centimeters or about 113 cubic centimeters. This page explains why by comparing the two rings slice by slice at matching heights, and finds the volume of every such ring from its height alone.

Key terms

Napkin ring problem

Definition Start with a solid ball of radius RR and drill a straight cylindrical hole of radius bb all the way through it, with 0<b<R0 < b < R and with the axis of the hole passing through the center of the ball. The solid that remains is a napkin ring. Its height hh is the distance, measured along the axis, from the bottom edge of the ring to its top edge, which is also the height of the straight inner wall left by the drill. In a vertical cut through the axis, the radius from the center of the ball to a point on the top edge has length RR, horizontal part bb and vertical part h/2h/2, so the Pythagorean theorem gives

(h2)2=R2−b2.\left(\frac{h}{2}\right)^2 = R^2 - b^2 .

The napkin ring problem asks for the volume of the ring. The answer is

V=πh36,V = \frac{\pi h^3}{6},

which depends on the height alone. So any two napkin rings of the same height have the same volume, whatever the sizes of the balls they were cut from, and this volume equals that of a solid ball of diameter hh. The claim is about volume only. Rings of the same height can have very different surface areas, and they have equal masses only when they are made of materials with the same density.

In plain words A bigger ball needs a wider hole to leave a ring of the same height. The bigger ring is wider around but has a thinner wall, and in the volume these two effects cancel exactly.

Example A ball of radius 55 with a hole of radius 44 leaves a ring of height 225−16=62\sqrt{25 - 16} = 6. A ball of radius 13\sqrt{13} with a hole of radius 22 leaves a ring of height 213−4=62\sqrt{13 - 4} = 6 as well. Both rings therefore have volume

π⋅636=36π≈113.1\frac{\pi \cdot 6^3}{6} = 36\pi \approx 113.1

cubic units, the volume of a solid ball of radius 33. Their surfaces differ. The outer band of a sphere of radius RR between two parallel planes a distance hh apart has area 2πRh2\pi R h, and the wall of the hole has area 2πbh2\pi b h, so the first ring has surface area 108π108\pi and the second 12(13+2)π≈67.3π12(\sqrt{13} + 2)\pi \approx 67.3\pi square units.

Cross-section

Definition Stand a solid upright along a chosen vertical axis and measure height zz along that axis. The cross-section of the solid at height zz is the flat region in which the horizontal plane at height zz meets the solid. Its area A(z)A(z) is the cross-sectional area at that height, and A(z)=0A(z) = 0 at heights where the plane misses the solid.

In plain words Cut the solid straight across, the way a loaf of bread is sliced, and look at the face of the cut. That face is the cross-section, and its area says how much of the solid lies at that level.

Example Center a ball of radius 55 at height 00. The plane at height 33 meets it in a disk. A point on the rim of that disk is at distance 55 from the center of the ball and 33 above it, so the disk has radius 25−9=4\sqrt{25 - 9} = 4 and area 16π16\pi. The same reasoning shows that a ball of radius rr centered at height 00 has, at each height zz with −r≤z≤r-r \le z \le r, a disk cross-section of area

A(z)=π(r2−z2).A(z) = \pi \left(r^2 - z^2\right).

Annulus

Definition An annulus is the flat region between two circles with a common center, an outer circle of radius routr_{\text{out}} and an inner circle of radius rinr_{\text{in}}, where 0≤rin≤rout0 \le r_{\text{in}} \le r_{\text{out}}. Its area is the area of the outer disk minus the area of the inner disk,

A=πrout2−πrin2=π(rout2−rin2).\begin{aligned} A &= \pi r_{\text{out}}^2 - \pi r_{\text{in}}^2 \\[0.4em] &= \pi \left(r_{\text{out}}^2 - r_{\text{in}}^2\right). \end{aligned}

When the inner radius is 00 the annulus is a whole disk, and when the two radii are equal its area is 00.

In plain words An annulus is a flat ring, shaped like a washer. A larger outer circle does not by itself mean a larger area, because the hole in the middle may have grown too.

Example Every horizontal cross-section of a napkin ring is an annulus. Center the ring at height 00. For −h/2≤z≤h/2-h/2 \le z \le h/2, the plane at height zz cuts the original ball in a disk of radius R2−z2\sqrt{R^2 - z^2}, and the hole removes a disk of radius bb from its middle, so

A(z)=π(R2−z2−b2)=π(h24−z2),\begin{aligned} A(z) &= \pi \left(R^2 - z^2 - b^2\right) \\[0.4em] &= \pi \left(\frac{h^2}{4} - z^2\right), \end{aligned}

since R2−b2=h2/4R^2 - b^2 = h^2/4. The radius of the ball has disappeared from the result. For the ring cut from the ball of radius 55, the plane at height 22 gives outer radius 21\sqrt{21} and area π(21−16)=5π\pi(21 - 16) = 5\pi. For the ring cut from the ball of radius 13\sqrt{13} it gives outer radius 33 and area π(9−4)=5π\pi(9 - 4) = 5\pi, the same value, just as the formula π(9−z2)\pi(9 - z^2) for h=6h = 6 predicts. A plane above height h/2h/2 or below −h/2-h/2 meets no part of the ring, because there the slice of the ball lies entirely inside the hole or misses the ball.

Cavalieri’s principle

Definition Suppose two solids lie between the same two parallel planes, and every plane parallel to those two cuts both solids in cross-sections of equal area. Then the two solids have equal volumes. Only the areas have to match, not the shapes of the cross-sections. The principle holds for the solids of ordinary geometry, whose volumes and cross-sectional areas are defined, and in calculus it follows from the slicing formula

V=∫cdA(z) dzV = \int_{c}^{d} A(z)\, dz

for a solid that lies between heights cc and dd. Equal volumes do not imply equal cross-sections, so the principle cannot be run backward. It is named after Bonaventura Cavalieri, who stated it in a book of 1635.

In plain words Think of each solid as a stack of very thin slices. If the slices at every level have the same area in both stacks, the stacks hold the same amount, even when one is made of wide thin rings and the other of small disks. A neat pile of coins and the same pile pushed into a leaning tower have the same volume for the same reason.

Example Center three solids at height 00, the ring cut from the ball of radius 55, the ring cut from the ball of radius 13\sqrt{13} and a solid ball of radius 33. All three lie between the planes at heights −3-3 and 33. At each height zz in that range the two rings have annular cross-sections and the ball has a disk, and by the formulas of the previous two cards all three areas equal π(9−z2)\pi(9 - z^2). Cavalieri’s principle then says that the three solids have the same volume.

Volume of a sphere

Definition A sphere of radius rr is the surface made of all points at distance rr from a center. The volume of a sphere means the volume of the solid ball it encloses, which is

V=43πr3=πd36,V = \frac{4}{3}\pi r^3 = \frac{\pi d^3}{6},

where d=2rd = 2r is the diameter. Cavalieri’s principle gives this formula. At each height zz from 00 to rr, the upper half of a ball centered at height 00 has a disk cross-section of area π(r2−z2)\pi(r^2 - z^2). Next to it, stand a cylinder of radius rr between heights 00 and rr, and remove a cone whose tip is the center of the cylinder’s bottom face and whose base is the cylinder’s top face. At height zz what remains is an annulus with outer radius rr and inner radius zz, of the same area π(r2−z2)\pi(r^2 - z^2). So the half ball has the volume of the cylinder minus the cone,

πr3−πr33=2πr33,\pi r^3 - \frac{\pi r^3}{3} = \frac{2\pi r^3}{3},

and the whole ball has twice that.

In plain words A ball fills a little more than half of the smallest cube that holds it. The cube of side dd has volume d3d^3 and the ball has volume πd3/6\pi d^3/6, which is about 0.524d30.524 d^3.

Example A ball of radius 33 has volume 43π⋅27=36π\tfrac43 \pi \cdot 27 = 36\pi. For the napkin ring itself, a ring of height hh has cross-sectional area π(h2/4−z2)\pi(h^2/4 - z^2) at each height between −h/2-h/2 and h/2h/2, which is exactly the cross-sectional area of a solid ball of radius h/2h/2 centered at height 00. By Cavalieri’s principle the ring has the volume of that ball,

43π(h2)3=πh36.\frac{4}{3}\pi \left(\frac{h}{2}\right)^3 = \frac{\pi h^3}{6}.

With h=6h = 6 this gives 36π36\pi again.

Check yourself

  1. A ball of radius 1010 has a straight hole of radius 88 drilled through its center. Find the height of the napkin ring and its volume, and compare the area of its central cross-section with that of the solid ball of the same volume.
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Half the height is 100−64=6\sqrt{100 - 64} = 6, so the ring has height h=12h = 12. Its volume is

π⋅1236=288π≈904.8\frac{\pi \cdot 12^3}{6} = 288\pi \approx 904.8

cubic units, which is the volume 43π⋅63=288π\tfrac43 \pi \cdot 6^3 = 288\pi of a solid ball of radius 66. At the center the cross-section of the ring is an annulus with radii 1010 and 88, of area π(100−64)=36π\pi(100 - 64) = 36\pi. The central disk of the ball of radius 66 has area 36π36\pi as well, as the slice comparison requires.

  1. Two balls of radius 33 and 55 each get a straight hole of radius 22 drilled through the center. Do the two rings have the same volume? What does the answer say about the condition in the napkin ring problem?
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They do not. The ring from the smaller ball has half-height 9−4=5\sqrt{9 - 4} = \sqrt{5}, so its height is 25≈4.472\sqrt{5} \approx 4.47. The ring from the larger ball has half-height 25−4=21\sqrt{25 - 4} = \sqrt{21} and height 221≈9.172\sqrt{21} \approx 9.17. Their volumes are

V3=π(25)36=2053 π≈14.9π\begin{aligned} V_3 &= \frac{\pi \left(2\sqrt{5}\right)^3}{6} \\[0.4em] &= \frac{20\sqrt{5}}{3}\,\pi \approx 14.9\pi \end{aligned}

and

V5=π(221)36=2821 π≈128.3π.\begin{aligned} V_5 &= \frac{\pi \left(2\sqrt{21}\right)^3}{6} \\[0.4em] &= 28\sqrt{21}\,\pi \approx 128.3\pi . \end{aligned}

The formula V=πh3/6V = \pi h^3/6 still holds for each ring, but the heights differ, so the volumes differ. The surprise needs the heights to be equal. When the hole radius is held fixed instead, a bigger ball leaves a taller ring and therefore more material.

  1. A ball of radius 10001000 centimeters is drilled through its center so that the ring left over is 66 centimeters tall. Find the radius of the hole, the thickness of the ring at its middle, the area of its cross-section 22 centimeters above the center, and its volume.
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The hole radius bb satisfies

b2=10002−32=999991,b^2 = 1000^2 - 3^2 = 999991,

so b≈999.9955b \approx 999.9955 centimeters. At its middle the ring reaches from radius bb out to radius 10001000, a thickness of about 0.00450.0045 centimeters, thinner than a sheet of paper. With R=1000R = 1000 and R2−b2=9R^2 - b^2 = 9, the cross-section two centimeters above the center has area

A(2)=π(R2−b2−22)=π(9−4)=5π\begin{aligned} A(2) &= \pi\left(R^2 - b^2 - 2^2\right) \\[0.4em] &= \pi (9 - 4) = 5\pi \end{aligned}

square centimeters, the same as for the small rings of height 66. The volume is π⋅63/6=36π\pi \cdot 6^3/6 = 36\pi, about 113.1113.1 cubic centimeters, the same as for every other ring of height 66 centimeters.