Volume of Sphere from Surface Area

Theorem

The volume $V$ of a sphere of radius $r$ is given by:

$V = \dfrac {r A} 3$

where $A$ is the surface area of the sphere.

Let the surface of the sphere of radius $r$ be divided into many small areas.

If they are made small enough, they can be approximated to plane figures.

Let the areas of these plane figures be denoted:

$a_1, a_2, a_3, \ldots$

Let the sphere of radius $r$ be divided into as many pyramids whose apices are at the center and whose bases are these areas.

$\dfrac {r a_1} 3, \dfrac {r a_2} 3, \dfrac {r a_3} 3, \ldots$

The volume $\VV$ of the sphere is given by the sum of the volumes of each of these pyramids:

$\ds \VV$

$=$

$\ds \dfrac {r a_1} 3 + \dfrac {r a_2} 3 + \dfrac {r a_3} 3 + \cdots$

$\ds $

$=$

$\ds \dfrac r 3 \paren {a_1 + a_2 + a_3 + \cdots}$

But $a_1 + a_2 + a_3 + \cdots$ is the surface area of the sphere.

Hence:

$\ds \VV$

$=$

$\ds \dfrac r 3 \paren {a_1 + a_2 + a_3 + \cdots}$

$\ds $

$=$

$\ds \dfrac r 3 A$

It needs to be noted that this proof is intuitive and non-rigorous.

$\blacksquare$

This was the method used by Johannes Kepler as an offshoot of the proof he gave for the area of a circle

1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.10$: Kepler ($\text {1571}$ – $\text {1630}$)