Advanced view of the napkin ring problem? - MathOverflow

Advanced view of the napkin ring problem?

2010-08-31T19:01:02Z

The "napkin-ring problem" sometimes shows up in 2nd-year calculus courses, but it can fit quite neatly into a high-school geometry course via Cavalieri's principle.

However, the conclusion remains astonishing. Is there some advanced viewpoint from which it becomes obvious from some sort of symmetry that's not visible in the naive formulation?

Answer by David Eppstein for Advanced view of the napkin ring problem?

2010-08-31T21:04:17Z

I think it's a bit more elegant to use a different version of Cavalieri's principle: instead of taking cross-sections, sweep the napkin ring by half-disks, bounded by semicircles on the outer sphere whose diameter is the same as the napkin ring's height. The shape of the half-disk is independent of the outer radius. The instantaneous volume swept by the half-disk, per unit angle, just looks (in the limit as the angle goes to zero) as a wedge of a sphere, so it's also independent from the outer radius.

Edited to add: I think this is the argument in the following reference from the Wikipedia article: Levi, Mark (2009), "6.3 How Much Gold Is in a Wedding Ring?", The Mathematical Mechanic: Using Physical Reasoning to Solve Problems, Princeton University Press, pp. 102–104, ISBN 978-0-691-14020-9.

Answer by Steve Huntsman for Advanced view of the napkin ring problem?

2010-08-31T21:09:34Z

Let $V(r,z)$ denote the volume of a napkin ring of outer radius $r$ and height $2z$. We have that $V(r,rz) = r^3 V(1,z)$ and $V(1,az) = a^3 V(1,z)$. The former equality is trivial. To see the latter one, note that $V(1,z) = 4\pi z^3/3$.*

It follows that $V(r,z) = r^3 V(1,r^{-1}z) = r^3 r^{-3} V(1,z) = V(1,z)$.

*This might want some fleshing out: e.g. decomposing the complementary part of the sphere into two spherical cones, a cylinder, and two "negative" cones. The volumes of the cylinder and cones can be trivially computed; the volume of a spherical cone can be computed without calculus using its solid angle $\Omega$ and taking the proportion $\Omega/4\pi$.

Answer by Joseph O'Rourke for Advanced view of the napkin ring problem?

2010-09-01T01:53:21Z

This is my attempt to understand David Eppstein's construction. The green segment is the "stick connecting the center of the sphere to a point halfway from top to bottom on the inner surface of the hole." The half-disk that gets angularly swept around the center of the sphere is purple. The napkin ring ("hole") is red.

Edit. Altered as per David's comment.

Answer by Aaron Meyerowitz for Advanced view of the napkin ring problem?

2010-09-01T07:44:59Z

EDIT: More briefly than before . Here is a naive physical argument which might meet the request of a point of view from which the result would immediately appear to be just what one would expect before going through the argument.

A blob of (incompressible) fluid volume V will form a spherical ball of radius (what it needs to be) if uncontrained
A blob of fluid volume V constrained between two parallel plates at z=r and z=-r will form the height 2r central slice of a sphere of radius R where R is just right so that the slice has volume V (provided r is not too large relative to V in which case we get a sphere)
A blob of fluid of volume V constrained between two parallel plates at z=r and z=-r and with a cylinder of height 2r and radius q imposed in the middle will form (along with the cylinder) the height 2r central slice of a sphere of radius R where R is just right so that the slice has volume $V+\pi q^2 h$. This shape might not have the curved boundry reach the cylinder
Imagine that the volume is just right to get that napkin ring. Now start shrinking q. We will still have a napkin ring. Keep going until q=0 and we see that the volume was that of a sphere of radius r.

(read the previous version if you wish, it might not be worth it)

Answer by sleepless in beantown for Advanced view of the napkin ring problem?

2010-09-01T11:07:05Z

The key in seeing this lies in breaking it down into a series of stacked washers or discs.

Define the napkin ring of height $h$ as being hollowed out of a sphere of radius centered at the origin with radius $R=\alpha h$, with $\alpha>=1$
Define the hole as being drilled along the $z$-axis, creating the napkin ring with an outer diameter $r_{out}$ and inner diameter $r_{in}$ defined as functions of $z$.
$r_{out}=\sqrt(R^2-z^2)$ and inner diameter $r_{in}=\sqrt(R^2-(h/2)^2)$
Define the volume as the integral over $z$ ranging over ($-h/2, +h/2$)

$$\pi \int_{-h/2}^{h/2} (r_{out}^2-r_{in}^2) dz$$

as the volume being height ($dz$) multiplied by the area ($\pi r_{out}^2 - \pi r_{in}^2)$ of the outer disk minus the inner disk. Note that the $R^2$ cancels out, showing that the radius of the sphere of material does not affect the AREA of the annulus at height $z$.

This is the intuitive step that may be hard to grasp. The thickness of the napkin ring ($r_{out}-r_{in}$) certainly does change with the radius $R$ of the carving sphere. As $R$ increases, the thickness of the napkin ring decreases inversely proportional to the square of $R$. This inverse relationship is what keeps the areas of the annuli at height $z$ at a value that is independent of $R$.

The intuitive mis-step is in thinking that even as $R$ increases, the thickness of the napkin ring stays the same. This is incorrect. A similar intuitive mis-step also occurs with this example.

You are wearing a belt of circumference 1 meter. You let out the belt ~31.416 centimeters (~ ten belt notches). How far over your body will the belt float? (also, how much larger can you now become?) The answer is 10 centimeter increase in radius.

The earth is wearing a belt at its equator of circumference $C$. How much do you have to let the belt out to create a an increase in radius of 10 centimeters for the earth? The answer is the same ~31.416 centimeters, since circumference is linearly proportional to radius. $C=2\pi r$

In the napkin ring problem, the key is that the area of the annuslus as a function of distance along the longitudinal-axis remains constant despite any change in $R$ because of the inverse-square proportionality. Intuition fools us into thinking that the thickness must remain constant for such a thing.

Answer by Ashu Prakash for Advanced view of the napkin ring problem?

2011-04-03T17:00:50Z

Volume of a napkin ring = (4/3)h^3