Pattern · a volume that ignores its sphere

The Ring That Forgets Its Sphere

Drill a hole straight through the middle of a solid sphere and keep the ring that's left. Its volume is π·h³ / 6 — where h is the ring's height — and that is the whole answer. Not the sphere's radius. Not the drill's width. A ring cut from a grapefruit and a ring of the same height cut from a sphere the size of the Sun enclose exactly the same volume.

Here is the puzzle Martin Gardner used to set it. A cylindrical hole six centimetres long is drilled straight through the centre of a solid sphere. What volume of the sphere is left? It looks like you're missing a number — nobody told you how big the sphere was. You're not missing anything. The height of the hole is the only number you need, and everything else cancels. Below you can watch it cancel.

Slice it, and see why

The reason lives in the cross-sections. Take the leftover ring, and take a plain solid sphere whose diameter equals the ring's height h. Slice both at the same height. The ring gives a ring-shaped slice — an annulus; the little sphere gives a solid disk. Their areas are equal. At every height. Drag the knife and the sphere and watch.

The annulus is wide and thin; the disk is small and solid. They look nothing alike — and they have the same area, because the ring's outer radius √(R²−z²) and its hole a both grow with the sphere, and the difference of their squares, (R²−z²) − a², loses every trace of R. It equals (h/2)² − z² — the little sphere's slice — no matter how large the sphere you started with. Two solids with matching slices at every level have matching volume: that is Cavalieri's principle (1635), and it is the whole proof.

Now make the sphere enormous

Fix the ring's height at h = 5 cm and grow the sphere it was drilled from — from something you could hold up to something the size of a small mountain. The ring's shape lurches from a fat barrel to a razor-thin band. Its volume never moves. Watch the live slice-census below: it sums the ring's own annular slices, computed the honest way — π(R²−z²) − πa², with the sphere's radius appearing in every single term — and the total lands on the same value however large the sphere.