# geometry – Why is the amount of a cone one third of the amount of a cylinder?

*by*Phil Tadros

Here’s a derivation of the amount of a cone which doesn’t use calculus, Cavalieri’s precept, the tactic of exhaustion, or another infinitesimal arguments.

[**Edit**There is a flaw in this argument, see below] [

**Edit 2**The flaw has been fixed, by considering the ratio of the volume of a cone to its circumscribing cylinder under different scalings]

We will cut up a cone horizontally into two items, in order that the higher half types one other cone with a smaller base, and the decrease half is not a cone however an object referred to as a ‘frustum’.

For a cone with base radius $r$ and top $h$, we are able to use a parameter $b$ with $0 lt b lt 1$ to outline the peak of the frustum as $b h$.

As a result of the entire cone and the higher cone type comparable triangles within the vertical cross part, the higher cone with top $ (1-b) h $ could have a base radius of $ (1 – b) r $.

The amount of the frustum will likely be equal to the amount of the unique cone, much less the amount of the higher cone. We do not but know what type the operate representing the amount of a cone will take, so for now we are going to simply write $V_{cone} = V_{cone}(r,h)$ to remind us that it will likely be some operate of the peak and base radius. So the amount of the frustum is $$V_{frustum} = V_{cone}(r,h) – V_{cone}((1 – b)r,(1 – b)h)$$

At this level we make the remark that the *ratio* of the amount of a cone to the amount of it is circumscribing cylinder have to be invariant below a scaling on the coordinates (the ratio is homogeneous of diploma 0).

$$frac{V_{cone}(r,h)}{pi r^2 h} = frac{V_{cone}(sr,sh)}{pi (sr)^2 sh}$$

for all $s>0$. If we write $V_{cone} = hat{Q},F(r,h), r^2 h$ the place $F(r,h)$ is a few as but unknown operate and $hat{Q}$ is a continuing, then

$$F(r,h) = F(sr,sh)$$

so $F(r,h)$ can also be homogeneous of diploma 0.

Therefore

start{array}{l@{}l}

V_{frustum}

&{}= V_{cone}(r,h) – V_{cone}((1 – b)r,(1 – b)h)

&{}= hat{Q} , F(r,h) ,r^2 h – hat{Q} , F((1-b)r, (1-b)h) , (1-b)^2r^2 (1-b)h

&{}= hat{Q} , F(r,h) , r^2 h ( 1 – (1-b)^3)

&{}= Q , r^2 h (3 b – 3 b ^2 +b^3)

finish{array}

the place $Q = hat{Q} , F(r,h)$

Now take into account the next determine

It’s clear that the amount of the frustum of top $b h$ have to be *larger* than the interior cylinder of radius $ (1-b) r$ and top $b h$ and it should even be *much less* than the amount of the outer cylinder with radius $ r $ and top $b h$.

$$pi (1-b)^2r^2 b h lt V_{frustum} lt pi r^2 b h$$

Substituting the expression for $V_{frustum}$ from above and dividing every little thing via by $ b pi r^2 h $

$$ (1-b)^2 lt frac{Q (3 – 3b + b^2)}{pi} lt 1$$

This should maintain for all $0 lt b lt 1$.

At this level, we might use the acquainted argument about limits – particularly, as $b$ will get nearer to zero, the decrease certain approaches the higher certain of $1$, so $frac{Q 3}{pi} = 1$ or $Q = frac{pi}{3}$.

Nonetheless, it’s attainable to seek out the worth of $Q$ another way, that doesn’t contain some restrict course of.

First, observe that the worth of $Q$ has bounds positioned on it by the geometry of the issue $0 lt Q lt pi$ for the reason that cone should have some quantity, and that quantity have to be lower than the amount of a cylinder with radius $r$ and top $h$. What we’re going to present is that for all values of $Q$ on this vary, with only one exception, there’s a alternative of $b$ with $0 lt b lt 1$ that causes the above inequality to not maintain. Within the spirit of Sherlock Holmes, ‘..when you could have eradicated the unimaginable, no matter stays, nonetheless inconceivable [or in our case, expected], have to be the reality’.

We cut up the issue up into two elements. The higher certain of the inequality doesn’t maintain when

$$frac{Q (3 – 3b + b^2)}{pi} = 1$$

Fixing for $b$

$$b = frac{3}{2}-sqrt{frac{pi}{Q}-frac{3}{4}}$$

Now introduce a parameter $alpha$ and write $Q= pi / (1+alpha+alpha^2) $. Then for $0 lt alpha lt 1$ we now have $pi/3 lt Qltpi$ and the above equation reduces to $b=1-alpha$, so $0 lt b lt 1$.

The decrease certain of the inequality doesn’t maintain when

$$(1-b)^2 = frac{Q (3 – 3b + b^2)}{pi}$$

Fixing for $b$

$$b = 1 – frac{ (frac{1}{2} + sqrt{frac{pi}{Q}-frac{3}{4}})}{frac{pi}{Q}-1}$$

Introduce a parameter $alpha$ as earlier than, however this time write $Q=pi alpha^2 / (1+alpha+alpha^2) $. Then for $0 lt alpha lt 1$ we now have $0 lt Q lt pi/3$ and the above equation once more reduces to $b=1-alpha$, so $0 lt b lt 1$.

Subsequently we now have $0 lt Q lt pi$ by the geometry of the issue, however every time $0 lt Q lt frac{pi}{3}$ or $frac{pi}{3} lt Q lt pi$ there exists not less than one worth for $b$ with $0 lt b lt 1$ for which the inequality doesn’t maintain. The one remaining risk on the interval $0 lt Q lt pi$ is $Q=frac{pi}{3}$ (for all $r,h > 0$), and so

$$ V_{cone} = frac{pi}{3} r^2 h$$