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.
Proof
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.
From Volume of Pyramid, their volumes are:
- $\dfrac {r a_1} 3, \dfrac {r a_2} 3, \dfrac {r a_3} 3, \ldots$
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.
$\blacksquare$
Historical Note
This was the method used by Johannes Kepler as an offshoot of the proof he gave for the area of a circle
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.10$: Kepler ($\text {1571}$ – $\text {1630}$)