Circles, cones and spheres

Section B.5 Circles, cones and spheres

Subsection B.5.1 Where Does the Formula for the Area of a Circle Come From?

Typically when we come across $\pi$ for the first time it is as the ratio of the circumference of a circle to its diameter

\begin{align*} \pi &= \frac{C}{d} = \frac{C}{2r} \end{align*}

Indeed this is typically the first definition we see of $\pi\text{.}$ It is easy to build an intuition that the area of the circle should be propotional to the square of its radius. For example we can draw the largest possible square inside the circle (an inscribed square) and the smallest possible square outside the circle (a circumscribed square):

The smaller square has side-length $\sqrt{2} r$ and the longer has side-length $2r\text{.}$ Hence

\begin{align*} 2 r^2 & \leq A \leq 4r^2 & \text{ or } 2 & \leq \frac{A}{r^2} \leq 4 \end{align*}

That is, the area of the circle is between 2 and 4 times the square of the radius. What is perhaps less obvious (if we had not been told this in school) is that the constant of propotionality for area is also $\pi\text{:}$

\begin{align*} \pi &= \frac{A}{r^2}. \end{align*}

We will show this using Archimedes’ proof. He makes use of these inscribed and circumscribed polygons to make better and better approximations of the circle. The steps of the proof are somewhat involved and the starting point is to rewrite the area of a circle as

\begin{align*} A &= \frac{1}{2} C r \end{align*}

where $C$ is (still) the circumference of the circle. This suggests that this area is the same as that of a triangle of height $r$ and base length $C$

\begin{align*} T &= \frac{1}{2} C r \end{align*}

Archimedes’ proof then demonstrates that indeed this triangle and the circle have the same area. It relies on a “proof by contradiction” — showing that $T \lt A$ and $T \gt A$ cannot be true and so the only possibility is that $A=T\text{.}$

We will first show that $T \lt A$ cannot happen. Construct an $n$-sided “inscribed” polygon as shown below:

Let $p_n$ be the inscribed polygon as shown.

We need 4 steps.

The area of $p_n$ is smaller than that of the circle — this follows since we can construct $p_n$ by cutting slices from the circle.
Let $E_n$ be the difference between the area of the circle and $p_n\text{:}$ $E_n = A - A(p_n)$ (see the left of the previous figure). By the previous point we know $E_n \gt 0\text{.}$ Now as we increase the number of sides, this difference becomes smaller. To be more precise, we will now show that

\begin{align*} E_{2n} & \leq \frac{1}{2} E_n. \end{align*}

The error $E_n$ is made up of $n$ “lobes”. In the centre-left of the previous figure we draw one such lobe and surround it by a rectangle of dimensions $a \times 2b$ — we could determine these more precisely using a little trigonometry, but it is not necessary.

This diagram shows the lobe is smaller than the rectangle of base $2b$ and height $a$ Since there are $n$ copies of the lobe, we have

\begin{align*} E_n & \leq n \times 2ab & \text{rewrite as } \frac{E_n}{2} & \leq nab \end{align*}

Now draw in the polygon $p_{2n}$ and consider the associated “error” $E_{2n}\text{.}$ If we focus on the two lobes shown then we see that the area of these two new lobes is equal to that of the old lobe (shown in centre-left) minus the area of the triangle with base $2b$ and height $a$ (drawn in purple). Since there are $n$ copies of this picture we have

\begin{align*} E_{2n} &= E_n - nab & \text{now use that $nab \geq E_n/2$}\\ & \leq E_n - \frac{E_n}{2} = \frac{E_n}{2} \end{align*}
The area of $p_n$ is smaller than $T\text{.}$ To see this decompose $p_n$ into $n$ isosceles triangles. Each of these has base shorter than $C/n\text{;}$ the straight line is shorter than the corresponding arc — though strictly speaking we should prove this. The height of each triangle is shorter than $r\text{.}$ Thus
\begin{align*} A(p_n) &= n \times \frac{1}{2} \text{(base)}\times \text{(height)}\\ & \leq n \times \frac{Cr}{2n} = T \end{align*}
If we assume that $T \lt A\text{,}$ then $A-T = d$ where $d$ is some positive number. However we know from point 2 that we can make $n$ large enough so that $E_n \lt d$ (each time we double $n$ we halve the error). But now we have a contradiction to step 3, since we have just shown that
\begin{align*} E_n = A-A(p_n) & \lt d = A-T & \text{which implies that}\\ A(p_n) & \gt T. \end{align*}

Thus we cannot have $T \lt A\text{.}$

If we now assume that $T \gt A$ we will get a similar contradiction by a similar construction. Now we use regular $n$-sided circumscribed polygons, $P_n\text{.}$

The proof can be broken into 4 similar steps.

The area of $P_n$ is greater than that of the circle — this follows since we can construct the circle by trimming the polygon $P_n\text{.}$
Let $E_n$ be the difference between the area of the polygon and the circle: $E_n = A(P_n)-A$ (see the left of the previous figure). By the previous point we know $E_n \gt 0\text{.}$ Now as we increase the number of sides, this difference becomes smaller. To be more precise, we will now show

\begin{align*} E_{2n} & \leq \frac{1}{2} E_n. \end{align*}

The error $E_n$ is made up of $n$ “lobes”. In the centre-left of the previous figure we draw one such lobe. Let $L_n$ denote the area of one of these lobes, so $E_n = nL_n\text{.}$ In the centre of the previous figure we have labelled this lobe carefully and also shown how it changes when we create the polygon $P_{2n}\text{.}$ In particular, the original lobe is bounded by the straight lines $\vec{ad}, \vec{af}$ and the arc $\widehat{fbd}\text{.}$ We create $P_{2n}$ from $P_n$ by cutting away the corner triangle $\triangle aec\text{.}$ Accordingly the lines $\vec{ec}$ and $\vec{ba}$ are orthogonal and the segments $|bc|=|cd|\text{.}$

By the construction of $P_{2n}$ from $P_n\text{,}$ we have

\begin{align*} 2L_{2n} &= L_n - A( \triangle aec) & \text{or equivalently } L_{2n} &= \frac{1}{2} L_n - A( \triangle abc) \end{align*}

And additionally

\begin{align*} L_{2n} & \leq A( \triangle bcd) \end{align*}

Now consider the triangle $\triangle abd$ (centre-right of the previous figure) and the two triangles within it $\triangle abc$ and $\triangle bcd\text{.}$ We know that $\vec{ab}$ and $\vec{cb}$ form a right-angle. Consequently $\vec{ac}$ is the hypotenuse of a right-angled triangle, so $|ac| \gt |bc| = |cd|\text{.}$ So now, the triangles $\triangle abc$ and $\triangle bcd$ have the same heights, but the base of $\vec{ac}$ is longer than $\vec{cd}\text{.}$ Hence the area of $\triangle abc$ is strictly larger than that of $\triangle bcd\text{.}$

Thus we have

\begin{align*} L_{2n} & \leq A(\triangle bcd) \lt A(\triangle abc) \end{align*}

But now we can write

\begin{align*} L_{2n} &= \frac{1}{2} L_n - A( \triangle abc) \lt \frac{1}{2} L_n - L_{2n} & \text{rearrange}\\ 2L_{2n} & \lt \frac{1}{2} L_n & \text{there are $n$ such lobes, so }\\ 2n L_{2n} & \lt \frac{n}{2} L_n & \text{since $E_n = n L_n$, we have}\\ E_{2n} & \lt \frac{1}{2} E_n & \text{which is what we wanted to show.} \end{align*}
The area of $P_n$ is greater than $T\text{.}$ To see this decompose $P_n$ into $n$ isosceles triangles. The height of each triangle is $r\text{,}$ while the base of each is longer than $C/n$ (this is a subtle point and its proof is equivalent to showing that $\tan \theta \gt \theta$). Thus
\begin{align*} A(P_n) &= n \times \frac{1}{2} \text{(base)}\times \text{(height)}\\ & \geq n \times \frac{Cr}{2n} = T \end{align*}
If we assume that $T \gt A\text{,}$ then $T-A = d$ where $d$ is some positive number. However we know from point 2 that we can make $n$ large enough so that $E_n \lt d$ (each time we double $n$ we halve the error). But now we have a contradiction to step 3, since we have just shown that
\begin{align*} E_n = A(P_n) - A & \lt d = T-A & \text{which implies that}\\ A(P_n) & \lt T. \end{align*}

Thus we cannot have $T \gt A\text{.}$ The only possibility that remains is that $T=A\text{.}$

Subsection B.5.2 Where Do These Volume Formulas Come From?

We can establish the volumes of cones and spheres from the formula for the volume of a cylinder and a little work with limits and some careful summations. We first need a few facts.

Every square number can be written as a sum of consecutive odd numbers. More precisely
\begin{align*} n^2 &= 1 + 3 + \cdots + (2n-1) \end{align*}
The sum of the first $n$ positive integers is $\frac{1}{2} n(n+1)\text{.}$ That is
\begin{align*} 1 + 2 +3 + \cdots +n &= \frac{1}{2}n(n+1) \end{align*}
The sum of the squares of the first $n$ positive integers is $\frac{1}{6} n(n+1)(2n+1)\text{.}$
\begin{align*} 1^2 + 2^2 +3^2 + \cdots + n^2 &= \frac{1}{6}n(n+1)(2n+1) \end{align*}

We will not give completely rigorous proofs of the above identities (since we are not going to assume that the reader knows mathematical induction), rather we will explain them using pictorial arguments. The first two of these we can explain by some quite simple pictures:

We see that we can decompose any square of unit-squares into a sequence of strips, each of which consists of an odd number of unit-squares. This is really just from the fact that

\begin{align*} n^2 - (n-1)^2 &= 2n-1 \end{align*}

Similarly, we can represent the sum of the first $n$ integers as a triangle of unit squares as shown. If we make a second copy of that triangle and arrange it as shown, it gives a rectangle of dimensions $n$ by $n+1\text{.}$ Hence the rectangle, being exactly twice the size of the original triangle, contains $n(n+1)$ unit squares.

The explanation of the last formula takes a little more work and a carefully constructed picture:

Let us break these pictures down step by step

Leftmost represents the sum of the squares of the first $n$ integers.
Centre — We recall from above that each square number can be written as a sum of consecutive odd numbers, which have been represented as coloured bands of unit-squares.
Make three copies of the sum and arrange them carefully as shown. The first and third copies are obvious, but the central copy is rearranged considerably; all bands of the same colour have the same length and have been arranged into rectangles as shown.

Putting everything from the three copies together creates a rectangle of dimensions $(2n+1)\times(1+2+3+\cdots+n)\text{.}$

We know (from above) that $1+2+3+\cdots+n = \frac{1}{2} n(n+1)$ and so

\begin{align*} (1^2+2^2+\cdots+n^2 ) &= \frac{1}{3} \times \frac{1}{2} n(n+1)(2n+1) \end{align*}

as required.

Now we can start to look at volumes. Let us start with the volume of a cone; consider the figure below. We bound the volume of the cone above and below by stacks of cylinders. The cross-sections of the cylinders and cone are also shown.

To obtain the bounds we will construct two stacks of $n$ cylinders, $C_1,C_2,\dots,C_n\text{.}$ Each cylinder has height $h/n$ and radius that varies with height. In particular, we define cylinder $C_k$ to have height $h/n$ and radius $k \times r/n\text{.}$ This radius was determined using similar triangles so that cylinder $C_n$ has radius $r\text{.}$ Now cylinder $C_k$ has volume

\begin{align*} V_k &= \pi \times \text{radius}^2 \times \text{height} = \pi \left( \frac{kr}{n} \right)^2 \cdot \frac{h}{n}\\ &= \frac{\pi r^2h}{n^3} k^2 \end{align*}

We obtain an upper bound by stacking cylinders $C_1,C_2,\dots,C_n$ as shown. This object has volume

\begin{align*} V &= V_1 + V_2 + \cdots + V_n\\ &= \frac{\pi r^2h}{n^3} \left(1^2 + 2^2 + 3^2 + \cdots + n^2 \right)\\ &= \frac{\pi r^2h}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \end{align*}

A similar lower bound is obtained by stacking cylinders $C_1,\dots,C_{n-1}$ which gives a volume of

\begin{align*} V &= V_1 + V_2 + \cdots+ V_{n-1}\\ &= \frac{\pi r^2h}{n^3} \left(1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \right)\\ &= \frac{\pi r^2h}{n^3} \cdot \frac{(n-1)(n)(2n-1)}{6} \end{align*}

Thus the true volume of the cylinder is bounded between

\begin{align*} \frac{\pi r^2h}{n^3} \cdot \frac{(n-1)(n)(2n-1)}{6} & \leq \text{correct volume} \leq \frac{\pi r^2h}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \end{align*}

We can now take the limit as the number of cylinders, $n\text{,}$ goes to infinity. The upper bound becomes

\begin{align*} \lim_{n \to \infty} \frac{\pi r^2h}{n^3} \frac{n(n+1)(2n+1)}{6} &= \frac{\pi r^2h}{6} \lim_{n\to\infty} \frac{n(n+1)(2n+1)}{n^3}\\ &= \frac{\pi r^2h}{6} \lim_{n\to\infty} \frac{(1+1/n)(2+1/n)}{1}\\ &= \frac{\pi r^2h}{6} \times 2\\ &= \frac{\pi r^2h}{3} \end{align*}

The other limit is identical, so by the squeeze theorem we have

\begin{align*} \text{Volume of cone } &= \frac{1}{3} \pi r^2h \end{align*}

Now the sphere — though we will do the analysis for a hemisphere of radius $R\text{.}$ Again we bound the volume above and below by stacks of cylinders. The cross-sections of the cylinders and cone are also shown.

To obtain the bounds we will construct two stacks of $n$ cylinders, $C_1,C_2,\dots,C_n\text{.}$ Each cylinder has height $R/n$ and radius that varies with its position in the stack. To describe the position, define

\begin{align*} y_k &= k \times \frac{R}{n} \end{align*}

That is, $y_k\text{,}$ is $k$ steps of distance $\frac{R}{n}$ from the top of the hemisphere. Then we set the $k^\mathrm{th}$ cylinder, $C_k$ to have height $R/n$ and radius $r_k$ given by

\begin{align*} r_k^2 &= R^2 - (R-y_k)^2 = R^2 - R^2(1-k/n)^2\\ &= R^2( 2k/n - k^2/n^2) \end{align*}

as shown in the top-right and bottom-left illustrations. The volume of $C_k$ is then

\begin{align*} V_k &= \pi \times \text{radius}^2 \times \text{height} = \pi \times R^2 \left(2k/n - k^2/n^2 \right) \times \frac{R}{n}\\ &= \pi R^3 \cdot \left( \frac{2k}{n^2} - \frac{k^2}{n^3} \right) \end{align*}

We obtain an upper bound by stacking cylinders $C_1,C_2,\dots,C_n$ as shown. This object has volume

\begin{align*} V &= V_1 + V_2 + \cdots + V_n\\ &= \pi R^3 \cdot \left( \frac{2}{n^2}\left(1 + 2 + 3 + \cdots + n \right) - \frac{1}{n^3}\left(1^2 + 2^2 + 3^2 + \cdots + n^2 \right) \right) \end{align*}

Now recall from above that

\begin{align*} 1 + 2 + 3 + \cdots +n &= \frac{1}{2} n(n+1) & 1^2 + 2^2 + 3^2 + \cdots +n^2 &= \frac{1}{6} n(n+1)(2n+1) \end{align*}

\begin{align*} V &= \pi R^3 \cdot \left( \frac{n(n+1)}{n^2} - \frac{n(n+1)(2n+1)}{6n^3} \right) \end{align*}

Again, a lower bound is obtained by stacking cylinders $C_1,\dots,C_{n-1}$ and a similar analysis gives

\begin{align*} V &= \pi R^3 \cdot \left( \frac{n(n-1)}{(n-1)^2} - \frac{n(n-1)(2n-1)}{6(n-1)^3} \right) \end{align*}

Thus the true volume of the hemisphere is bounded between

\begin{align*} \pi R^3 \cdot \left( \frac{n(n+1)}{n^2} - \frac{n(n+1)(2n+1)}{6n^3} \right) & \leq \text{correct volume} \leq \pi R^3 \cdot \left( \frac{n(n+1)}{n^2} - \frac{n(n+1)(2n+1)}{6n^3} \right) \end{align*}

We can now take the limit as the number of cylinders, $n\text{,}$ goes to infinity. The upper bound becomes

\begin{align*} \lim_{n \to \infty} \pi R^3 \cdot \left( \frac{n(n+1)}{n^2} - \frac{n(n+1)(2n+1)}{6n^3} \right) &= \pi R^3 \left( \lim_{n\to\infty} \frac{n(n+1)}{n^2} - \frac{n(n+1)(2n+1)}{6n^3} \right)\\ &= \pi R^3 \left( 1 - \frac{2}{6} \right) = \frac{2}{3} \pi R^3. \end{align*}

The other limit is identical, so by the squeeze theorem we have

\begin{align*} \text{Volume of hemisphere } &= \frac{2}{3} \pi R^3 & \text{ and so}\\ \text{Volume of sphere } &= \frac{4}{3} \pi R^3 \end{align*}

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