#### What is the volume of a sphere?

Archimedes discovered, and proved, that a sphere occupies precisely two-thirds of the cylinder that contains it.

If the cylinder has a radius, r, the area of the base is πr2. As the cylinder has a height equal to the sphere, its volume must be πr2 × 2r = 2πr3, and so...

Theorem
The volume of the sphere, V, is given by V = 4πr3/3, where r is the radius.

Proof
Let us consider the top half of a sphere, radius r.

We shall say that the radius of the k th layer is Rk and if we split the height of the hemisphere into n layers, the k th layer will have a vertical height of Hk = (k/n)r.

Using the Pythagorean Theorem,
Rk2 = r2 – Hk2 = r2 (k/n)2r2 = r2(1 – k2/n2)

Therefore Rk2 = (r2/n2) (n2k2)

The area of the base of the k th layer will be πRk2 = (πr2/n2)(n2k2).

The height of each slab will be r/n, so the volume of the k th slab (assuming it to be a cylinder) will be (πr2/n2)(n2k2) × r/n = (πr3/n3)(n2k2).

Therefore, if the volume of the sphere is V, the volume of the hemisphere, ½V, can be approximated.

 V ≈ (πr3/n3)(n2 – 12) + (πr3/n3)(n2 – 22) + ... + (πr3/n3)(n2 – n2) = (πr3/n3)((n2 – 12) + (n2 – 22) + ... + (n2 – n2) = (πr3/n3)(n × n2 – (12 + 22 + ... + n2 )) = (πr3/n3)(n3 – (12 + 22 + ... + n2 ))

Using the result
12 + 22 + 32 + ... + n2 = 1/6 n(n + 1)(2n + 1)

 V ≈ (πr3/n3)(n3 – 1/6 n(n + 1)(2n + 1)) = (πr3/6)(6 – (1 + 1/n)(2 + 1/n))

As n tends towards infinity, 1/n tends towards zero and so ½V tends towards (πr3/6)(6 – 2) = 2πr3/3.

Hence the volume of a sphere, V = 4πr3/3.