Napkin Ring Problem - Proof

Proof

Suppose the radius of the sphere is and the length of the cylinder (or the tunnel) is .

By the Pythagorean theorem, the radius of the cylinder is

and the radius of the horizontal cross-section of the sphere at height y above the "equator" is

The cross-section of the band with the plane at height y is the region inside the larger circle of radius given by (2) and outside the smaller circle of radius given by (1). The cross-section's area is therefore the area of the larger circle minus the area of the smaller circle:

\begin{align} & {}\quad \pi(\text{larger radius})^2 - \pi(\text{smaller radius})^2 \\ & = \pi\left(\sqrt{R^2 - y^2}\right)^2 - \pi\left(\sqrt{R^2 - \left(\frac{h}{2}\right)^2\,{}}\,\right)^2 = \pi\left(\left(\frac{h}{2}\right)^2 - y^2\right). \end{align}

The radius R does not appear in the last quantity. Therefore the area of the horizontal cross-section at height y does not depend on R. The volume of the band is

and that does not depend on R.

This is an application of Cavalieri's principle: volumes with equal-sized corresponding cross-sections are equal. Indeed, the area of the cross-section is the same as that of the corresponding cross-section of a sphere of radius h/2, which has volume

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