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# Volume of a cone.
The volume of a cone with base area $B$ and height $H$ is given by $\displaystyle\frac{1}{3}BH$, but why?
![[1 teaching/summer program 2023/puzzles-and-problems/---files/volume-of-a-cone 2023-08-16 14.29.42.excalidraw.svg]]
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When compared with a cylinder of the same base area $B$ and height $H$, which has volume $BH$, we see that the volume of a cone ought to be some fraction of that of the corresponding cylinder, where $$
\text{volume of cone} \propto BH
$$or, $$
\text{volume of cone } = c BH
$$for some proportionality constant $c$.
We would like to figure out why $\displaystyle c = \frac{1}{3}$.
===
First, let us take a cone of base area $B$ and height $H$, and let us cut it horizontally somewhere to create a **frustum** and a **tiny cone**: ![[1 teaching/summer program 2023/puzzles-and-problems/---files/volume-of-a-cone 2023-08-16 14.22.02.excalidraw.svg]]
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Here the frustum has height $H_{1}$, and a top area of $b$ and bottom area of $B$, while the tiny cone has a height $H_{2}$, and a base area of $b$.
Since the tiny cone is a cone, by our proposal it has volume $cbH_{2}$.
The frustum therefore has volume $$
\text{vol. of frustum} =cBH-cbH_{2}
$$
This is a correct formula for the frustum, but it is a bit odd to use, because the frustum is described in terms of the variables $H_{1}, B$, and $b$, yet the expression we have has $H$ and $H_{2}$.
Let us see if we can re-write the volume of the frustum in terms of its height $H_{1}$ and its two base areas, $B$ and $b$ instead.
We notice that the big cone and the small cone are proportional to each other. If we denote $R$ as the radius of the base area $B$ of the big cone, and $r$ as the radius of the base area $b$ of the small cone,
![[1 teaching/summer program 2023/puzzles-and-problems/---files/volume-of-a-cone 2023-08-16 14.38.45.excalidraw.svg]]
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we see that we have proportion $$
\frac{H}{R} = \frac{H_{2}}{r}
$$
Since $B=\pi R^{2}$ and $b=\pi r^{2}$, we have $\displaystyle R=\sqrt{\frac{B}{\pi}}$ and $\displaystyle r=\sqrt{\frac{b}{\pi}}$. Substituting this into above gives $$
\frac{H}{\sqrt{\frac{B}{\pi}}} = \frac{H_{2}}{\sqrt{\frac{b}{\pi}}}
$$and simplifying gives $$
\frac{H}{\sqrt{B}} = \frac{H_{2}}{\sqrt{b}}.
$$
And since $H=H_{1}+H_{2}$, we have $$
\frac{H_{1}+H_{2}}{\sqrt{B}}=\frac{H_{2}}{\sqrt{b}}
$$
We like to focus on the frustum, so we will express the heights in terms of $H_{1}$. Using above, we can solve $H_{2}$ in terms of $H_{1}$ as $$
\begin{align*}
& \sqrt{b}H_{1}+\sqrt{b}H_{2}=H_{2}\sqrt{B} \\
\implies H_{2} &= \frac{\sqrt{b}}{\sqrt{B}-\sqrt{b}}H_{1}
\end{align*}
$$
And we can also solve for $H$ in terms of $H_{1}$, since $H=H_{1}+H_{2}$ we get $$
\begin{align*}
H &=H_{1} + \frac{\sqrt{b}}{\sqrt{B}-\sqrt{b}}H_{1} \\
\implies H &= \frac{\sqrt{B}}{\sqrt{B}-\sqrt{b}}H_{1}
\end{align*}
$$ So the volume of the frustum is $$
\begin{align*}
\text{vol. of frustum} &= cBH-cbH_{2} \\
& =cB \frac{\sqrt{B}}{\sqrt{B}-\sqrt{b}}H_{1}-cb \frac{\sqrt{b}}{\sqrt{B}-\sqrt{b}}H_{1} \\
& =cH_{1}\left( \frac{B\sqrt{B}}{\sqrt{B}-\sqrt{b}} - \frac{b\sqrt{b}}{\sqrt{B}-\sqrt{b}} \right) \\
\implies\color{blue}\text{vol. of frustum} &\color{blue}= cH_{1} \frac{B\sqrt{B}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}
\end{align*}
$$
Ok, that's kind of neat, we get the volume of the frustum in terms of its height $H_{1}$ and two base areas $b$ and $B$. But we still don't know what the proportionality constant $c$ is.
Now we use a limiting idea. What if we let the area $b\to B$? Then the frustum becomes a cylinder of height $H_{1}$ and base area $B$ !
So the volume should converge to the volume of this cylinder, namely $$
\lim_{b\to B} cH_{1} \frac{B \sqrt{B}- b \sqrt{b}}{\sqrt{B}-\sqrt{b}} = H_{1}B
$$
We can compute the limit of the left side. Either we do **L'Hospital rule** or **resolve the algebra**, we get the limit on the left side to become $$
\lim_{b\to B} cH_{1} \frac{B \sqrt{B}- b \sqrt{b}}{\sqrt{B}-\sqrt{b}} = 3cH_{1}B
$$
And since we have equality $$
3cH_{1} B = H_{1}B
$$we find $3c=1$, or $\displaystyle c=\frac{1}{3}$!
In other words, the volume of a cone with base area $B$ and height $H$ is $$
\frac{1}{3}BH.
$$
===
**Questions 1.**
In this derivation, we needed to take the limit $$
\lim_{b\to B} cH_{1} \frac{B \sqrt{B}- b \sqrt{b}}{\sqrt{B}-\sqrt{b}}
$$
which we see is in $\frac{0}{0}$ indeterminant form. Solve this limit in two different ways:
(A) Using the conjugate $\sqrt{B}+\sqrt{b}$ and working out the algebra, to get a new expression that the limit is easier to calculate.
(B) Use L'Hospital rule to compute this limit.
In each case you should get the same (and expected) answer.
**Question 2.**
The following five limits look similar, but they are **all different**. Be careful and calculate them (either with algebra or L'Hospital rule, if needed at all.)
(A) $\displaystyle \lim_{b\to B} \frac{B\sqrt{B}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}$
(B) $\displaystyle \lim_{b\to B} \frac{B\sqrt{b}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}$
(C) $\displaystyle \lim_{b\to B} \frac{b\sqrt{B}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}$
(D) $\displaystyle \lim_{b\to B} \frac{b\sqrt{b}-b\sqrt{b}}{\sqrt{B}-\sqrt{b}}$
(E) $\displaystyle \lim_{b\to B} \frac{b\sqrt{B}-B\sqrt{b}}{\sqrt{B}-\sqrt{b}}$