# 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 $V = \frac {h\pi}{3} \left( R^2 + Rr + r^2 \right)$ - 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:

$$\pi \int_0^h \left(\frac{R-r}{h}x + r \right)^2 dx$$

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.

$$\pi \int_0^h \left(kx + r \right)^2 dx = \pi \int_0^h k^2x^2 + 2krx + r^2 dx \\= \pi \left[\frac13 k^2x^3 + kr x^2 + r^2 x \right]_0^h = \pi \left[\frac13 k^2h^3 + kr h^2 + r^2 h \right]$$

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 $k$:

$$V = \pi \left[\frac13 \frac{(R-r)^2}{h^2}h^3 + \frac{R-r}{h} r h^2 + r^2 h \right]$$

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

$$V = \pi \left[\frac13 (R-r)^2h + (R-r)r h + r^2 h \right]$$

Factor out the $h$ and expand the brackets:

$$V = h\pi \left[\frac13 (R^2- 2rR + r^2) + Rr - r^2 + r^2 \right]$$

We're almost there! Obviously the last two terms cancel each other out; I'm also going to take the $\frac13$ out of the bracket to give me:

$$V = \frac{h\pi}{3} \left[ R^2- 2rR + r^2 + 3Rr \right]$$

$$V = \frac{h\pi}{3} \left[ R^2 +rR + r^2 \right]$$

How reassuring!

In practise, you can probably ignore the $\pi$ and the $3$ ($\frac \pi3$ is about $1.05$, so unless you're doing things very accurately, it'll be noise.

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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