Volumes of Solid of Revolution - Watermelon ft. Archimedes,Newton and Leibniz

A watermelon has an ellipsoidal shape with major axis 28 cm and minor axis 25 cm. We have to Find its volume.

Historical Approach: Before calculus, one way of approximating the volume would be to slice the watermelon (say in 2 cm thick slices) and add up the volumes of each slice using $\displaystyle{V}=\pi{r}^{2}{h}$V=πr​2​​h.

Interestingly, Archimedes (the one who famously jumped out of his bath and ran down the street shouting "Eureka! I've got it") used this approach to find volumes of spheres around 200 BC. The technique was almost forgotten until the early 1700s when calculus was developed by Newton and Leibniz.

We see how to do the problem using both approaches.

Volume using historical method:

Because the melon is symmetrical, we can work out the volume of one half of the melon, and then double our answer.
The radii for the slices for one half of a particular watermelon are found from measurement to be:
$\displaystyle{0},{6.4},{8.7},$0,6.4,8.7, $\displaystyle{10.3},{11.3},$10.3,11.3, $\displaystyle{12.0},{12.4},{12.5}.$12.0,12.4,12.5.
The approximate volume for one half of the melon using slices 2 cm thick would be:

$\displaystyle\text{V}_{{\text{half}}}=\pi\times{\left[{6.4}^{2}+{8.7}^{2}+{10.3}^{2}+{11.3}^{2}+{12.0}^{2}+{12.4}^{2}+{12.5}^{2}\right]}\times{2}$V​half​​=π×[6.4​2​​+8.7​2​​+10.3​2​​+11.3​2​​+12.0​2​​+12.4​2​​+12.5​2​​]×2

$\displaystyle=\pi\times{8040.44}\times{2}$=π×8040.44×2

$\displaystyle={5054.4}$=5054.4

So the volume for the whole watermelon is about
$\displaystyle{5054.4}\times{2}={10109}\ \text{cm}^{3}={10.1}\ \text{L}$5054.4×2=10109 cm​3​​=10.1 L.
In the following section, we see how to find the "exact" value using the volume of solid of revolution formula.

 "Exact" Volume (using Integration):

We are told the melon is an ellipsoid. We need to find the equation of the cross-sectional ellipse with major axis 28 cm and minor axis 25 cm.
We use the formula (from the section on ellipses):

$\displaystyle\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={1}$​a​2​​​​x​2​​​​+​b​2​​​​y​2​​​​=1

where a is half the length of the major axis and b is half the length of the minor axis.
For the volume formula, we will need the expression for y² and it is easier to solve for this now (before substituting our a and b).

$\displaystyle\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}={1}$​a​2​​​​x​2​​​​+​b​2​​​​y​2​​​​=1
$\displaystyle{b}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}$b​2​​x​2​​+a​2​​y​2​​=a​2​​b​2​​
$\displaystyle{a}^{2}{y}^{2}={a}^{2}{b}^{2}-{b}^{2}{x}^{2}={b}^{2}{\left({a}^{2}-{x}^{2}\right)}$a​2​​y​2​​=a​2​​b​2​​−b​2​​x​2​​=b​2​​(a​2​​−x​2​​)
$\displaystyle{y}^{2}=\frac{{{b}^{2}}}{{{a}^{2}}}{\left({a}^{2}-{x}^{2}\right)}$y​2​​=​a​2​​​​b​2​​​​(a​2​​−x​2​​)

Since $\displaystyle{a}={14}$a=14 and $\displaystyle{b}={12.5}$b=12.5, we have:

$\displaystyle{y}^{2}=\frac{{{12.5}^{2}}}{{{14}^{2}}}{\left({14}^{2}-{x}^{2}\right)}$y​2​​=​14​2​​​​12.5​2​​​​(14​2​​−x​2​​)

$\displaystyle={0.797}{\left({196}-{x}^{2}\right)}$=0.797(196−x​2​​)

NOTE: The a and b that we are using for the ellipse formula are not the same a and b we use in the integration step. They are completely different parts of the problem.

Using this, we can now find the volume using integration. (Once again we find the volume for half and then double it at the end).

$\displaystyle\text{V}_{{\text{half}}}=\pi{\int_{{0}}^{{14}}}{y}^{2}{\left.{d}{x}\right.}$V​half​​=π∫​0​14​​y​2​​dx

$\displaystyle=\pi{\int_{{0}}^{{14}}}{0.797}{\left({196}-{x}^{2}\right)}{\left.{d}{x}\right.}$=π∫​0​14​​0.797(196−x​2​​)dx

$\displaystyle={0.797}\pi{\int_{{0}}^{{14}}}{\left({196}-{x}^{2}\right)}{\left.{d}{x}\right.}$=0.797π∫​0​14​​(196−x​2​​)dx

$\displaystyle={2.504}{{\left[{196}{x}-\frac{{{x}^{3}}}{{{3}}}\right]}_{{0}}^{{14}}}$=2.504[196x−​3​​x​3​​​​]​0​14​​

$\displaystyle={2.504}{\left[{196}{\left({14}\right)}-\frac{{{14}^{3}}}{{{3}}}\right]}$=2.504[196(14)−​3​​14​3​​​​]

$\displaystyle={2.504}\times{1829.33}$=2.504×1829.33

$\displaystyle={4580.65}\ \text{cm}^{3}$=4580.65 cm​3​​

So the watermelon's total volume is $\displaystyle{2}\times{4580.65}={9161}\ \text{cm}^{3}$2×4580.65=9161 cm​3​​ or $\displaystyle{9.161}\ \text{L}$9.161 L. This is about the same as what we got by slicing the watermelon and adding the volume of the slices.