Maths · sample calcsheet · 3 pages

The napkin ring problem

Drill a cylindrical hole through the centre of a sphere and a band is left, of height h. Its volume turns out not to depend on the sphere at all — only on h. Symbolic maths shows why.

Both figures are by Michael Hardy, public domain, via the Wikipedia article on the napkin ring problem.

First page of the The napkin ring problem sample calcsheet in Maths Explore — The first page of the calcsheet as it appears in Maths Explore.

What this calculation covers

The volume of the band

The band is the slice of the sphere between the two cuts, less the cylinder drilled out of it. Both are standard volumes, and the hole's radius follows from Pythagoras: r² = R² − (h/2)².

Slice of the sphere between the cuts
Cylinder drilled out, r² substituted
Volume of the band itself

Where the sphere goes

With R and h left as symbols there is nothing to work out numerically, so the answer comes back as a formula — and R is not in it.

The band's volume, whatever sphere it came from
Rate of change with the sphere's radius: exactly zero

Two very different spheres

Height of the band
A small sphere
A much larger one
Radius of the hole
Hole in the small sphere
Hole in the large one
Band from the small sphere
Band from the large one
The two bands are equal
Both match the closed form

Results summary

A 50 mm sphere and a 200 mm sphere, each drilled to leave a 60 mm band, hold the same 113 097 mm³ — though one hole is 40 mm across and the other nearly 198 mm. The band is thick and stubby in the small sphere, thin and tall in the large one, and the two effects cancel exactly.

Try changing…

Set R₁ := 30 mm — the hole closes up and the band becomes a whole 60 mm sphere. The volume still does not move.
Take the band to 120 mm, with R₁ := 100 mm to hold it — eight times the volume, because it goes as the cube of the height.
Right-click the simplify row and pick Show as Function Call — the same answer, the other spelling.

All samples