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
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
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
We obtain an upper bound by stacking cylinders \(C_1,C_2,\dots,C_n\) as shown. This object has volume
A similar lower bound is obtained by stacking cylinders \(C_1,\dots,C_{n-1}\) which gives a volume of
Thus the true volume of the cylinder is bounded between
We can now take the limit as the number of cylinders, \(n\text{,}\) goes to infinity. The upper bound becomes
The other limit is identical, so by the squeeze theorem we have
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
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
as shown in the top-right and bottom-left illustrations. The volume of \(C_k\) is then
We obtain an upper bound by stacking cylinders \(C_1,C_2,\dots,C_n\) as shown. This object has volume
Now recall from above that
so
Again, a lower bound is obtained by stacking cylinders \(C_1,\dots,C_{n-1}\) and a similar analysis gives
Thus the true volume of the hemisphere is bounded between
We can now take the limit as the number of cylinders, \(n\text{,}\) goes to infinity. The upper bound becomes
The other limit is identical, so by the squeeze theorem we have