Frequently Asked Questions
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...
The volume of the sphere, V, is given by V = 4πr3/3, where r is the radius.
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) (n2 k2)
The area of the base of the k th layer will be πRk2 = (πr2/n2)(n2 k2).
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)(n2 k2) × r/n = (πr3/n3)(n2 k2).
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.