Maths in the real world: flowerpot volumes

A reader asks: how do I figure out the volume of soil I need to fill a flowerpot?

A flowerpot is a slightly peculiar shape: it’s not a cone, it’s not a cylinder, it’s somewhere between the two. Luckily, we have a word for such shapes: it’s a frustum of a cone.

The easiest thing to do, of course, is to look up the volume - but unless you know what the shape is called, you’ll struggle to find that the volume of a flowerpot is $V=\frac{h\pi }{3}(R^2+Rr+r^2)$ - where $R$ is the radius at one end, $r$ the radius at the other, and $h$ the vertical height.

If you didn’t have an internet handy, you could always turn to the calculus. A flowerpot has a handy axis of rotation - it’s symmetrical about the axis you’d plant a flower through. That means you can use the volume of revolution technique to work it out for yourself.

2-4-6-8, what are we going to integrate? Square the $y$! Times by $\pi$!

You can make a mathematical flowerpot by rotating a line around an axis. If the line starts on the $x$-axis, you get a cone; if you truncate the line a bit higher up, you get a frustum. In our case, we’re going to let the line go through the points $(0,r)$ and $(h,R)$, so the circles at the end are the right size.

A little bit of juggling gives us the equation of the line we want; its gradient is $\frac{R-r}{h}$ and it goes through $(0,r)$, so the equation is $y=\frac{R-r}{h}x+r$.

We need to work out $\pi {\int }_0^h{y}^{2}dx$, which is the formula for volumes of revolution (it’s the limit of the sum of the volumes of infinitesimally thin discs of radius $y$, and thickness $dx$, since you ask.) That gives us the monster:

$$\begin{gathered} pi \\ int\mathrm{\_}{0}^h \\ left( \\ fracR-rhx+r \\ right{)}^{2}dx \end{gathered}$$

Now, that’s going to be a bit tedious, even if it boils down to the integral of $(kx+c{)}^2$. I’m going to let $k=\frac{R-r}{h}$ for the moment, just to save on typing.

$$\begin{gathered} pi \\ int\mathrm{\_}{0}^h \\ left(kx+r \\ right{)}^{2}dx= \\ pi \\ int\mathrm{\_}{0}^h{k}^2{x}^2+2krx+r^{2}dx \\ = \\ pi \\ left\text{[} \\ frac13k^2{x}^3+kr{x}^2+r^{2}x \\ right \end{gathered}$$\_0^h = \pi \left$$\begin{gathered} frac13k^2{h}^3+kr{h}^2+r^{2}h \\ right \end{gathered}$$\]

Phew! Still not very nice-looking, but we’re getting there. It’s all algebra rom here. Let’s put the fraction back in for $k$:

$$\begin{gathered} V= \\ pi \\ left\text{[} \\ frac13 \\ frac(R-r{)}^2{h}^2{h}^3+ \\ fracR-rhr{h}^2+r^{2}h \\ right \end{gathered}$$\]

… and notice that a lot of the $h$s cancel:

$$\begin{gathered} V= \\ pi \\ left\text{[} \\ frac13(R-r{)}^{2}h+(R-r)rh+r^{2}h \\ right \end{gathered}$$\]

Factor out the $h$ and expand the brackets:

$$\begin{gathered} V=h \\ pi \\ left\text{[} \\ frac13(R^2-2rR+r^2)+Rr-r^2+r^2 \\ right \end{gathered}$$\]

We’re almost there! Obviously the last two terms cancel each other out; I’m also going to take the $\frac{1}{3}$ out of the bracket to give me:

$$\begin{gathered} V= \\ frach \\ pi3 \\ left\text{[}{R}^2-2rR+r^2+3Rr \\ right \end{gathered}$$\] $$\begin{gathered} V= \\ frach \\ pi3 \\ left\text{[}{R}^2+rR+r^2 \\ right \end{gathered}$$\]

How reassuring!

In practise, you can probably ignore the $\pi$ and the $3$ ($\frac{\pi }{3}$ is about $1.05$, so unless you’re doing things very accurately, it’ll be noise.