Spheres

One length decides it

A sphere is a perfectly round ball. Every point on its surface is the same distance from its center, and that distance is its radius, r. A line straight through the center from one side to the other is a diameter, 2r long.

Every sphere is the same shape; spheres differ only in size, and the radius fixes the size. So the volume of a sphere depends on its radius and nothing else.

The radius r runs from the center to the surface, and it is the same length in every direction.

Inside a snug cylinder

Fit the sphere snugly inside a cylinder, so that it touches the top, the bottom and the curved side. The cylinder’s radius is r, the same as the sphere’s, and its height is the sphere’s diameter, 2r. A cylinder’s volume is its circular cross section times its height, so this one holds $\pi r^2\times 2r=2\pi r^3$πr² × 2r = 2πr³.

The sphere fills exactly $\frac{2}{3}$2/3 of that cylinder. Archimedes proved this more than two thousand years ago by cutting the solids into thin slices and comparing the slices; that proof is a later lesson of its own.

The sphere touches the cylinder’s top, bottom and curved side. The cylinder’s radius is r and its height is 2r, the sphere’s diameter.

Why $\frac{2}{3}$2/3 is believable

The sphere takes up less room than the cylinder: the space between the ball and the cylinder’s rims, near the top and near the bottom, is empty.

It takes up more room than two cones inside it. Stand one cone on the sphere’s middle circle, with its apex at the top of the sphere, and hang another from the same circle, with its apex at the bottom. Each cone has radius r and height r, so each holds $\frac{1}{3}\times \pi r^2\times r=\frac{1}{3}\pi r^3$⅓ × πr² × r = ⅓πr³, and the two together hold $\frac{2}{3}\pi r^3$⅔πr³. That is a third of the cylinder’s $2\pi r^3$2πr³. The cones have straight sides, and the sphere bulges out beyond them.

So the sphere holds more than a third of the cylinder and less than all of it, and $\frac{2}{3}$2/3 lies between. Archimedes put it another way: a cone, a sphere and a cylinder of the same width and height hold amounts in the ratio 1 : 2 : 3.

The volume of a sphere

The cylinder holds $2\pi r^3$2πr³, and the sphere is $\frac{2}{3}$2/3 of it: $\frac{2}{3}\times 2\pi r^3=\frac{4}{3}\pi r^3$⅔ × 2πr³ = 4/3 πr³. So the volume of a sphere of radius r is $V=\frac{4}{3}\pi r^3$V = 4/3 πr³.

A sphere with a radius of 3 cm has $r^3=3\times 3\times 3=27$r³ = 3 × 3 × 3 = 27, and $\frac{4}{3}\times 27=36$4/3 × 27 = 36, so its volume is $36\pi$36π cm³, which is about 113.1 cm³.

A ball 10 cm across has a diameter of 10 cm, so its radius is 5 cm. Then $r^3=125$r³ = 125, and its volume is $\frac{4}{3}\times \pi \times 125\approx 523.6$4/3 × π × 125 ≈ 523.6 cm³.

A hemisphere is half a sphere, so its volume is half of $\frac{4}{3}\pi r^3$4/3 πr³, which is $\frac{2}{3}\pi r^3$⅔πr³.

Three slips

The radius is cubed, because a volume multiplies three lengths. Squaring it instead gives $\frac{4}{3}\times \pi \times 9=12\pi$4/3 × π × 9 = 12π for a radius of 3 cm, only a third of the right answer, $36\pi$36π.

Dropping the ÷ 3 from $\frac{4}{3}$4/3 gives $4\pi r^3$4πr³, which is three times too much: $108\pi$108π for a radius of 3 cm.

The formula uses the radius. Putting in the diameter, 6 cm instead of 3 cm, gives $6^3=216$6³ = 216 in place of 27, which is 8 times too much.

The surface of a sphere

Painting a sphere needs the area of its surface, not its volume. That area is $4\pi r^2$4πr², four times the area of a circle with the same radius. Archimedes found this too, and its proof also comes later. An area has $r^2$r², in square units such as $m^2$m²; a volume has $r^3$r³, in cubic units such as $m^3$m³. Checking the power of r keeps the two formulas apart.

A cubic meter holds 1000 liters: a liter is 1000 cm³, and a cubic meter is 100 × 100 × 100 = 1,000,000 cm³.