# In the spirit of Archimedes’ method of exhaustion

Shown in Fig. 1 is a pile of $n$ cylinders that approximates one half of a sphere.

Fig. 1

From Fig. 1, we see $r_i^2 = r^2-h_i^2$.

If $\Delta V_i$ denotes the volume of $i^{th}$ cylinder in the pile, then

$\Delta V_i = \pi \cdot r_i^2 \cdot \frac{r}{n}$

$= \pi (r^2-h_i^2) \cdot \frac{r}{n}$

$= \pi (r^2-(i\cdot\frac{r}{n})^2)\cdot \frac{r}{n}$

$= \frac{\pi r^3}{n} (1-(\frac{i}{n})^2)$.

Let

$V_{*}=\sum\limits_{i=1}^{n}\Delta V_i$,

we have

$V_{*}= \sum\limits_{i=1}^{n}\frac{\pi r^3}{n} (1-(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}\sum\limits_{i=1}^{n}(1-(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}(\sum\limits_{i=1}^{n}1-\sum\limits_{i=1}^{n}(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}(n-\frac{1}{n^2}\sum\limits_{i=1}^{n}i^2)$

$= \frac{\pi r^3}{n}(n-\frac{n(n+1)(2n+1)}{6n^2})\quad\quad\quad($see “Little Bird and a Recursive Generator$)$

$= \pi r^3(1-\frac{(n+1)(2n+1)}{6n^2})$

$= \pi r^3(1-\frac{2n^2+3n+1}{6n^2})$

$= \pi r^3(1-\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6})$.

As $n \rightarrow \infty,$

$V_* \rightarrow \pi r^3 (1-\frac{1}{3}) = \frac{2}{3}\pi r^3=V_{\frac{1}{2}sphere}.$

Therefore, $V_{sphere}= 2\cdot V_{\frac{1}{2}sphere}$ gives

$V_{sphere} = \frac{4}{3}\pi r^3\quad\quad\quad(1-1)$

Similarly, we approximate a cone

by a stack of $n$ cylinders:

Fig. 2

Since

$h_i = i\cdot\frac{h}{n}\quad\quad\quad(2-1)$

and

$\frac{h_i}{h} = \frac{r_i}{r}$,

we have

$r_i = \frac{r\cdot h_i}{h}\overset{(2-1)}{=}\frac{r}{h}\cdot i\frac{h}{n}=\frac{r\cdot i}{n}.\quad\quad\quad(2-2)$

If $\Delta V_i$ denotes the $i^{th}$ cylinder, then

$\Delta V_i = \pi\cdot r_i^2\cdot\frac{h}{n}\overset{(2-2)}{=}\frac{\pi r^2 i^2 h}{n^3}$.

Let

$V_* = \sum\limits_{i=1}^{n}\Delta V_i,$

we have

$V_*= \frac{\pi r^2 h}{n^3}\sum\limits_{i=1}^{n} i^2$

$= \frac{\pi r h}{n^3} \cdot \frac{2n^3+3n^2+n}{6}$

$= \pi r^2 h (\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2})$.

Since as $n \rightarrow \infty$, $V_*\rightarrow \frac{1}{3}\pi r^2 h$, we have

$V_{cone} = \frac{1}{3}\pi r^2 h.\quad\quad\quad(2-3)$

We now derive the formula for a sphere’s surface area.

Let us approximate a sphere by many cones:

Fig. 3

From Fig. 3, we see that as $\Delta A_i \rightarrow 0$,

$h \rightarrow r, \;\sum\limits_{i}\Delta A_i \rightarrow A_{sphere}$.

Consequently,

$\sum\limits_{i} \Delta V_i \overset{(2-3)}{=} \sum\limits_{i} \frac{1}{3}\Delta A_i h = \frac{1}{3}h \sum\limits_{i}\Delta A_i \rightarrow \frac{1}{3}rA_{sphere}$

and,

$\sum\limits_{i} \Delta V_i \rightarrow V_{sphere} \overset{(1-1)}{=} \frac{4}{3}\pi r^3$.

It follows that

$\frac{1}{3}r A_{sphere} = \frac{4}{3}\pi r^3.$

Hence,

$A_{sphere} = 4 \pi r^2.$