The Napkin Ring Problem: Same Height, Same Volume

Drilling the hole: setting up the ring

Start with a solid sphere of radius R and bore a cylindrical hole of radius a straight through its center, the drill’s axis passing exactly through the middle. Everything inside the cylinder is removed. What remains is a bracelet-shaped solid — a napkin ring, named for its resemblance to the rings that hold table napkins.

Put the center at the origin with the drill axis along the z-axis. The sphere is x²+y²+z²=R², and the cylindrical bore is x²+y²=a². The ring’s flat top and bottom rims sit exactly where the cylinder wall meets the sphere: there x²+y²=a², so z²=R²−a² and z=±√(R²−a²).

The single most important quantity is the ring’s height h — the length of the drilled hole, measured rim to rim along the axis:

h = 2√(R² − a²),   equivalently   a² = R² − h²⁄4.

A larger sphere (bigger R) needs a wider bore (bigger a) to keep the same height. And there is a floor: we must have R ≥ h/2, since a sphere smaller than that cannot possibly yield a ring of height h. When R = h/2 exactly, the bore radius a shrinks to zero — the “ring” is a whole undrilled sphere of diameter h.

The astonishing result

Here is the theorem that makes the napkin ring famous: the ring’s volume depends only on its height h, and not in any way on the radius R of the sphere it was cut from. Explicitly,

V = πh³ ⁄ 6.

Two rings of the same height — one bored from a cherry, one from a bowling ball, one from a planet — enclose exactly the same volume, down to the last cubic millimeter. Moreover, πh³/6 is precisely the volume of a whole solid ball whose diameter equals h (radius h/2): (4/3)π(h/2)³ = πh³/6. So a napkin ring of height h holds exactly as much material as an undrilled marble of diameter h.

This is an equality, not an approximation — no limits, no small-quantity fudging. The degenerate check confirms it: set a=0 (no hole) and the “ring” becomes the whole sphere, whose diameter is then h=2R, giving πh³/6 = π(2R)³/6 = (4/3)πR³, the familiar volume of a ball.

Why it works: equal slices and Cavalieri’s principle

Slice the ring by horizontal planes perpendicular to the drill axis. At height z (with |z| ≤ h/2) the slice is a flat annulus — a washer. Its outer edge is the sphere, at radius √(R²−z²); its inner edge is the straight cylinder wall, at the constant radius a. The washer’s area is the big disk minus the punched-out disk:

A(z) = π(R² − z²) − πa² = π(R² − a² − z²) = π(h²⁄4 − z²).

Watch what just happened. The sphere radius R and the bore radius a appear only through the combination R²−a², which we already pinned to h²/4. The sphere’s radius has cancelled. Every slice’s area is fixed by h alone: a bigger sphere makes a larger outer disk, but the wider hole punches out exactly enough to compensate, slice by slice.

Now compare a solid ball of diameter h. Its slice at height z is a full disk of radius √((h/2)²−z²), whose area is π(h²/4−z²) — identical to the ring’s washer at the same height. Two solids whose horizontal cross-sections have equal area at every height must have equal volume: this is Cavalieri’s principle, published by Bonaventura Cavalieri in his 1635 Geometria indivisibilibus continuorum. The ring and the ball match one another slice for slice, so they must match in volume — no integral is even required.

Doing the calculus two ways

If you prefer to sum the slabs directly, integrate the washer areas from the bottom rim to the top:

V = ∫_−h/2^h/2 π(h²⁄4 − z²) dz.

The integrand is even, so V = 2π∫₀^h/2(h²/4 − z²)dz = 2π[(h²/4)z − z³/3]₀^h/2 = 2π(h³/8 − h³/24) = 2π·(h³/12) = πh³/6.

A cylindrical-shell computation confirms the answer from the other direction. A thin shell at radius r (with a ≤ r ≤ R) has height 2√(R²−r²), so

V = ∫_a^R 2πr · 2√(R²−r²) dr = (4π⁄3)(R²−a²)^3/2 = (4π⁄3)(h⁄2)³ = πh³⁄6.

Once more, R survives only inside the combination R²−a² = h²/4, and once more it cancels. Whichever way you sum the solid — horizontal washers or vertical shells — the sphere’s size evaporates from the final number.

Bulk versus bore: the intuition

The result feels impossible because intuition tracks only one of two competing effects. Enlarge the sphere and its raw volume grows like R³ — a great deal more material. But to keep the height fixed you must widen the hole, and the discarded cylinder-plus-caps grows at exactly the same rate. The excess bulk and the enlarged bore are not merely comparable; they cancel identically, leaving πh³/6 untouched. In the Earth example, the surviving band is thinner than a single atom, yet its ideal volume is still exactly 36π ≈ 113 cm³.

The cancellation is special to three dimensions and to circular cross-sections. The areas of both the sphere’s slice and the bore scale as (radius)², and it is the difference of squares R²−a² that h holds constant. Try the same trick in the plane — cut a straight slab out of a disk — and the cross-sections are lengths, √(R²−z²)−a, which stubbornly refuse to combine into anything R-free. The napkin ring is a genuinely three-dimensional coincidence, born from the fact that area is proportional to radius squared.

The six-inch hole and a little history

The theorem hides inside a famous puzzle: “A cylindrical hole six inches long is drilled straight through the center of a solid sphere. What volume of the sphere remains?” The problem looks underspecified — surely you need to know the sphere’s size. But the napkin ring theorem guarantees the answer is fixed by the 6-inch height alone, so you are free to take the easiest case: shrink the sphere until the bore radius is zero and the sphere’s diameter is exactly 6 inches. The “ring” is then a whole 6-inch ball, and V = π·6³/6 = 36π ≈ 113.1 cubic inches — the same answer for every sphere, from a marble to a moon.

The underlying equal-slice reasoning traces to Cavalieri (1635) and, in spirit, to Archimedes. His celebrated result that a sphere’s volume is two-thirds that of its circumscribing cylinder rests on the same comparison of cross-sections. The napkin-ring puzzle itself is a staple of recreational mathematics, popularized in the twentieth century by writers such as Martin Gardner and, more recently, by video explainers (Vsauce, Mathologer) that have carried it to millions of viewers.

Cousins, cautions, and generalizations

Archimedes’ hat-box theorem. A band (zone) of a sphere of radius R cut by two parallel planes a distance h apart has lateral surface area exactly 2πRh — equal to the surface of the same-height band of the enclosing cylinder. It is the surface-area analogue of the same slice-by-slice thinking, and it is what makes cylindrical map projections of the sphere equal-area. Note that here R does appear, unlike in the volume result.

Pappus’s centroid theorem. The volume of a solid of revolution equals the generating area times the distance travelled by its centroid. It gives a third, entirely different route to the ring’s volume and underlies a whole family of “spinning” volume problems.

What the result is not. The magic is not that “holes don’t matter.” A hole of a fixed radius removes very different volumes from different spheres; it is specifically the fixed height that locks the answer. Nor does the clean cancellation survive a change of dimension: in the plane the cross-sections are lengths rather than circular areas, so the two-dimensional analogue simply fails. The napkin ring is a small, exact miracle that lives precisely in three dimensions.

Same height, same volume: a 6 cm-tall napkin ring bored from spheres of wildly different sizes. The bore radius a is whatever it must be to keep the height at 6 cm; the volume never changes.

Sphere	Radius R	Bore radius a (for h = 6 cm)	Ring volume V
Cherry	4 cm	√7 ≈ 2.65 cm	36π ≈ 113.1 cm³
Orange	6 cm	√27 ≈ 5.20 cm	36π ≈ 113.1 cm³
Bowling ball	10.9 cm	≈ 10.48 cm	36π ≈ 113.1 cm³
Earth	6,371 km	≈ 6,371 km (wall thinner than an atom)	36π ≈ 113.1 cm³