Archimedes and the Volume of a Sphere

What is to be proved

The volume of a sphere of radius r is $\frac{4}{3}\pi r^3$4/3 πr³. Put another way, the sphere fills exactly $\frac{2}{3}$2/3 of the cylinder that fits snugly around it, with radius r and height 2r: that cylinder holds $\pi r^2\times 2r=2\pi r^3$πr² × 2r = 2πr³, and $\frac{2}{3}\times 2\pi r^3=\frac{4}{3}\pi r^3$⅔ × 2πr³ = 4/3 πr³.

The Greek mathematician Archimedes proved this in the third century BC, by cutting solids into thin slices and comparing the slices. He was so pleased with the result that he asked for a sphere inside a cylinder to be carved on his tomb.

The sphere touches the cylinder’s top, bottom and curved side. The cylinder holds $2\pi r^3$2πr³, and the sphere fills $\frac{2}{3}$2/3 of it.

Equal slices, equal volumes

The proof rests on one idea. Stand two solids of the same height side by side, and cut them both across at the same height. If the two cuts always have the same area, at every height, then the two solids have the same volume.

To see why, cut both solids into very thin layers, like a pile of coins. A thin layer is very nearly a prism, so it holds its area times its thickness. Each layer of one solid has the same area and the same thickness as the layer beside it, so the two layers hold the same amount. Layer by layer the two solids hold the same amount, so in total they do too. This is Cavalieri’s principle, named after the Italian mathematician Bonaventura Cavalieri, who stated it in 1635; Archimedes used the idea nearly two thousand years before him.

Archimedes compared a hemisphere, half a sphere, with a solid whose volume is already known. Stand the hemisphere of radius r on its flat face. Beside it stand a cylinder of the same radius r and the same height r. From the cylinder, drill out a cone whose apex is the center of the cylinder’s base and whose base is the cylinder’s top circle.

Left, the hemisphere. Right, the cylinder with the cone drilled out: the dashed lines are the cone, with its apex at the center of the base. Both are cut at the same height. Below them are the two cuts seen from above: a disc from the hemisphere and a ring from the drilled cylinder.

The slice of the hemisphere

Cut the hemisphere at height h above its base. The cut is a circle. Call its radius s. The center of the sphere is the center of the hemisphere’s flat face. The center of the cut is h straight above it, and the edge of the cut lies on the sphere, so every point of the edge is r from the center of the sphere.

The center of the sphere, the center of the cut and a point on the edge of the cut make a right-angled triangle. Its hypotenuse is r, and its other two sides are h and s. By Pythagoras’ theorem, $s^2+h^2=r^2$s² + h² = r², so $s^2=r^2-h^2$s² = r² − h². The area of the cut is $\pi s^2=\pi \left(r^2-h^2\right)$πs² = π(r² − h²).

Drag the cut up and down. At every height the disc cut from the hemisphere and the ring cut from the drilled cylinder have the same area, $\pi \left(r^2-h^2\right)$π(r² − h²).

The slice of the drilled cylinder

Cut the drilled cylinder at the same height h. Without the hole, the cut would be a circle of radius r. The cone takes a circle out of the middle, so the cut is a ring.

The cone’s side runs in a straight line from its apex, at the center of the base, to the rim of the top, which is r across and r up. So the cone widens exactly as fast as it rises, and at height h its radius is h. The ring is a circle of radius r with a circle of radius h taken out, so its area is $\pi r^2-\pi h^2=\pi \left(r^2-h^2\right)$πr² − πh² = π(r² − h²).

That is the area of the hemisphere’s cut at the same height. At h = 0 both are $\pi r^2$πr², a full circle; at h = r both are 0, the top of the hemisphere and the apex of the cone. At every height between, the two cuts have the same area, so by Cavalieri’s principle the hemisphere and the drilled cylinder have the same volume.

The cut through the drilled cylinder at height h: a ring, a circle of radius r with a circle of radius h taken out. Its area, $\pi r^2-\pi h^2$πr² − πh², equals the area of the hemisphere’s cut at the same height.

Cylinder minus cone

The drilled cylinder’s volume is easy to find. The cylinder has radius r and height r, so it holds $\pi r^2\times r=\pi r^3$πr² × r = πr³. The cone has the same base and the same height, so it holds a third of that, $\frac{1}{3}\pi r^3$⅓πr³. What is left is $\pi r^3-\frac{1}{3}\pi r^3=\frac{2}{3}\pi r^3$πr³ − ⅓πr³ = ⅔πr³.

So the hemisphere holds $\frac{2}{3}\pi r^3$⅔πr³, and the whole sphere, two hemispheres, holds $2\times \frac{2}{3}\pi r^3=\frac{4}{3}\pi r^3$2 × ⅔πr³ = 4/3 πr³.

The cone, the hemisphere and the cylinder all have radius r and height r, and they hold $\frac{1}{3}\pi r^3$⅓πr³, $\frac{2}{3}\pi r^3$⅔πr³ and $\pi r^3$πr³: in the ratio 1 : 2 : 3.

A sphere of radius 6 cm, for example, has $r^3=216$r³ = 216, and $\frac{4}{3}\times 216=288$4/3 × 216 = 288, so its volume is $288\pi$288π cm³, about 904.8 cm³.

Two slips

Writing the ring’s area as $\pi {\left(r-h\right)}^2$π(r − h)². That squares the width of the ring, but the ring is a whole circle with a whole circle taken out, $\pi r^2-\pi h^2$πr² − πh². With r = 5 and h = 3 the ring is $\pi \left(25-9\right)=16\pi$π(25 − 9) = 16π, while $\pi {\left(5-3\right)}^2$π(5 − 3)² is only $4\pi$4π.

Taking the hemisphere to be the cone that was drilled out, $\frac{1}{3}\pi r^3$⅓πr³, or the whole cylinder, $\pi r^3$πr³. The hemisphere matches what is left when the cone is taken away: $\frac{2}{3}\pi r^3$⅔πr³.