fbpx

Proof 7: Pyramid and cone volume

The Cheops pyramid in Egypt is constructed of many equally thick layers of box-shaped stones (more or less). Imagine more and more such layers being thinner and thinner, and you eventually get a smooth and close to perfect pyramid. So, the number of layers $n$ must be a very large number before we can think of it as a pyramid.

A pyramid’s volume consists of each of the $n$ prism-shaped layers’ volume:

$$\begin{align}
V_\text{pyramid}&=\underbrace{V_1+V_2+\cdots+V_{n_1}+V_n}_{n \text{ terms}}\\
~&=\underbrace{A_1h_\text{layer}+A_2 h_\text{layer} +\cdots+A_{n_1} h_\text{layer} +A_n h_\text{layer} }_{n \text{ terms}}\qquad\leftarrow \text{Prism volume, } V=A h_\text{layer} \\
~&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} h_\text{layer} \\
~&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} \frac 1n h \qquad \leftarrow \text{ Each layer is a bit of the full height, } h_\text{layer} =\frac 1n h
\end{align}$$

The $n$ layers’ heights all-together make up the pyramid height $h$. Each is an $n$’th of this full height, so $ h_\text{layer} =\frac 1n h $.

All lengths become smaller with one $n$’th at a time up through the layers. Lengths in the second-bottommost layer (which has number $n-1$) are then missing one $n$’th, so there are $1-\frac 1n=\frac{n-1}n$ left, those in the third layer from the bottom, layer no. $n-2$, are missing two $n$’ths $\frac 2n$ and only have $\frac{n-2}n$ left etc. Areas consist of some lengths times some lengths times, and with all such lengths reducing by an $n$’th in each step, the areas reduce by one $n$’th times one $n$’th:

$$
A_{n-1}=\frac{n-1}n \frac{n-1}n A_\text{base}= \left(\frac{n-1}n\right)^2 A_\text{base}= \frac{(n-1)^2}{n^2} A_\text{base}\\
A_{n-2}=\frac{n-2}n \frac{n-2}n A_\text{base}= \left(\frac{n-2}n\right)^2 A_\text{base}= \frac{(n-2)^2}{n^2} A_\text{base}\\
\vdots\\
A_2=\frac2n \frac2n A_\text{base}= \left(\frac2n\right)^2 A_\text{base}= \frac{2^2}{n^2} A_\text{base}\\
A_1=\frac1n \frac1n A_\text{base}= \left(\frac1n\right)^2 A_\text{base}= \frac{1^2}{n^2} A_\text{base} $$

And $A_n$ can of course be written as $A_n=\frac{n^2}{n^2} A_\text{base} $ to stay in pattern (dividing a number by itself gives 1; and it makes no difference to multiply with 1). Let’s put all these areas in:

$$\begin{align}
V_\text{pyramid}&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} \frac 1n h\\
~ &=\underbrace{\left( \frac{1^2}{n^2} A_\text{base} + \frac{2^2}{n^2} A_\text{base} +\cdots+ \frac{(n-1)^2}{n^2} A_\text{base} + \frac{n^2}{n^2} A_\text{base} \right) }_{n \text{ terms}} \frac 1n h\\
~ &=\underbrace{\left( \frac{1^2}{n^2} + \frac{2^2}{n^2} +\cdots+ \frac{(n-1)^2}{n^2} + \frac{n^2}{n^2} \right) }_{n \text{ terms}} \frac 1n h \,A_\text{base} \\
~ &=\underbrace{\left( 1^2 +2^2 +\cdots+ (n-1)^2 + n^2 \right) }_{n \text{ terms}} \frac 1{n^3} h \,A_\text{base}
\end{align}$$

From here we need a math trick. We can’t continue with this confusingly long row of terms (so long that we can’t write them all out but only can indicate them with the dots $\cdots$). Luckily a mathematician has proven that a sum of many squared numbers in a row actually has a formula:

$$1^2+2^2+\cdots+(n-1)^2+n^2=\frac16n(n+1)(2n+1)$$

This formula is called The sum of squares (see Proof 8). It may not be pretty but at least we can see all terms involved so we can calculate them. We plug it in instead of the long row:

$$\begin{align}
V_\text{pyramid}&= \underbrace{\frac16n(n+1)(2n+1)}_{\text{row of }n\text{ terms replaced}} \frac 1{n^3} h \,A_\text{base} \\
~ &= \frac16(n+1)(2n+1) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16(2n^2+n+2n+1^2) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16(2n^2+3n+1) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16\left(2+\frac3n+\frac1{n^2}\right) h \,A_\text{base} \\
~ &= \left(\frac13+\frac1{2n}+\frac1{6n^2}\right) h \,A_\text{base}
\end{align}$$

The more layers in the pyramid, the more perfect it becomes. We increase $n$ to be enormous – almost infinitely large, $n\to \infty$. When $n$ becomes enormous, the denominators in the fractions $ \frac1{2n}$ and $ \frac1{6n^2}$ become enormous, which makes the fractions insignificantly small. When $n$ is enormous it thus corresponds to these fractions completely disappearing, $ \frac1{2n}\to 0$ and $ \frac1{6n^2}\to 0$:

$$V_\text{pyramid}= \left(\frac13+\frac1{2n}+\frac1{6n^2}\right) h \,A_\text{base}\to \left(\frac13+0+0\right) h \,A_\text{base}\quad \text{when }n\to\infty$$

And finally, the result is a simple formula:

$$V_\text{pyramid}=\frac13 hA_\text{base}\quad_\blacksquare$$

The shape of the base didn’t matter along the way, so this formula counts for any shape, any number of edges – also for a round base, in other words for a cone:

$$V_\text{cone}=\frac13 h\underbrace{A_\text{base}}_\text{circle}=\frac13\pi r^2 h\quad_\blacksquare$$

Responses