# The napkin ring problem

I heard an interesting thought experiment on a podcast some time ago, but I am not sure of the origins of the problem. This problem is the *napkin ring problem*. Despite hearing this problem some time ago I have only just found time to convince myself that it is true, surprising as it is. I thought I would share the solution since it is a lovely application of multiple integrations. The problem is as follows:

Take a tennis ball and drill a cylindrical hole exactly through the centre of the ball so a to leave a ring around the circumference whose height is 2cm. (This ring would resemble a napkin ring.) Now take a much larger ball, say the Earth (if this were a perfect sphere), and perform the same task, leaving a ring around the equator of height 2cm. Then these two rings will have precisely the same volume.

On first hearing, this seems impossible. The Earth is much larger than a tennis ball. Taking even a small ring around the equator must be much larger than the tennis ball. Then we draw a few pictures - not to scale, of course - and see that as the sphere becomes larger, the size of the ring around the circumference gets much thinner.

If we fix a small length, say 1cm, then Pythagoras’s theorem tells us that the distance from the centre of the sphere to the inner edge of the ring is given by the square root of the radius squared minus this length squared. (Assuming the length is smaller than the radius, this can be done.) When the the length is relatively large compared to the radius, as in the case of the tennis ball, the thickness of the ring is also quite large. However, when the length is very small compared to the radius, the thickness of the ring must be very thin, as in the case of the Earth.

In case this is not clear, let’s put some numbers in to see how the thickness changes. Let’s suppose that the radius of the tennis ball is 3cm and the radius of the earth is 1000m - that is 100000cm, which is a gross underestimate of the radius of the earth. We take our small length to be 1cm. As in our argument above, the thickness of the ring for the tennis ball is 3cm minus the square root of 8, which gives approximately 0.172. Comparatively, performing the same calculation for the Earth gives the thickness of the ring as 0.000005cm. (For the actual radius of the Earth, this quantity is tiny!)

The argument above does not show that the volume does not change in these two cases. For this, we need to use a little calculus. What we actually need is to calculate the area of the segment of the circle that is “cut off” to make the ring. The volume is $2\pi$ multiplied by this area. I should warn you here, the argument get’s a little technical at this point.

To get the area of this segment, we use the following double integral
\[
A = \int_{-\ell}^{\ell}\int_{\sqrt{R^2 - \ell^2}}^{\sqrt{R^2 - z^2}} r ,\mathrm{d}r,\mathrm{d}z = 2.
\]
Here $R$ is the radius of the sphere, $\ell$ is the height of the ring measured from the diameter (we are assuming, for simplicity, that the sphere is centred at the origin), and $z$ is the vertical distance from the horizontal diameter. The quantity $r$ denotes the distance from the centre of the cylindrical drilled-out section to a point in the sphere. (Technically, we are using *cylindrical polar coordinates* to evaluate this integral.) The limits of integration are obtained using some careful applications of Pythagoras’s theorem. Evaluating the integral, we see that the area is two-thirds of the length $\ell$ cubed; in particular, this is independent of the radius $R$. Thus the volume of the ring is given by
\[
\frac{4\pi\ell^3}{3}=2.
\]
This is a lovely problem, where the mathematics gives a relatively surprising fact. The solution using multiple integrals does not necessarily provide any intuition as to why this is true, but it is a nice application of vector calculus to a fairly real-world problem.