This appendix reviews various formulae for finding the curved surface area and volume of a cone.

Surface area of a cone

Here’s a diagram of a cone. We’ve drawn some lengths and an angle on it.

A cone with the height, the base radius, the slant height and half the opening angle labelled.

When we cut the (slanted) surface of the cone in a straight line from the vertex to the base and then unwrap it, we end up with a sector of a circle, like this:

The cone, unwrapped.

The radius of this sector is the slant length of the cone, which is \(l\). The outer arc of the sector was the circumference of the base circle of the cone, so its length is \(2\pi r\).

Now the circumference of the whole of this circle is \(2\pi l\), and therefore the sector is a fraction \(\dfrac{2\pi r}{2\pi l}\) of the whole circle. The area of the sector, and hence of the slanted surface of the cone, is therefore \[\dfrac{2\pi r}{2\pi l}\times \pi l^2=\pi rl.\]

Alternatively, we could have worked out \(\phi\) first, using the formula for the arc length of a sector “\(l=r\theta\)”, giving in this case \(2\pi r=l\phi\), so \(\phi=\frac{2\pi r}{l}\). Then the area of the sector is given by the formula “\(A=\frac12 r^2\theta\)”, so calculating \(\frac12 l^2 .\frac{2\pi r}{l}=\pi rl\) gives us the area as before.

It’s always good to check that our answers are sensible before going too far in a problem!

How could we do this here?

Here are a few things that we can do to check that our answer is sensible.

  1. Try special values.

    What values could we put in to our variables that would give us something we can check?

    In this case, if we make our cone very “flat”, so the vertex is barely above the base, the cone will look very much like a circular disk of radius \(r\). Then the slant length \(l\) will be almost the same as \(r\), and the answer our formula gives us, \(\pi rs\) will be almost the same as \(\pi rr=\pi r^2\), which is the area of a circular disk of radius \(r\).

    This gives us some confidence that our answer is reasonable.

  2. Check the dimensions.

    The formula we have come up with, \(A=\pi rl\) (where \(A\) is the sloping surface area), must have correct dimensions to stand any chance of being correct.

    The left hand side is an area, measured in \(\quantity{}{m^2}\), so we say it has dimension \([\mathrm{L}^2]\), where \([\mathrm{L}]\) means a length.

    The right hand side is \(\pi\), a pure number (so dimensionless), times \(r\), which is a length, times \(l\), which is a length. So it has dimension \([\mathrm{L}]\times[\mathrm{L}]=[\mathrm{L}^2]\).

    So the dimensions of the left and right hand sides agree, giving us confidence that our answer is at least plausible.

Sometimes we will know some measurements more readily than others. So we could try rewriting this answer using other variables.

Since \(l^2=r^2+h^2\) using Pythagoras, we could also write the surface area as \(\pi r\sqrt{r^2+h^2}\) or \(\pi l\sqrt{l^2-h^2}\) if we wish.

If we draw a cross-section of the cone, we see that we have a right-angled triangle:

Cross-section of a cone.

From this diagram, we see that \(h=l\cos\theta\) and \(r=l\sin\theta\), so we can substitute for \(l\) in our earlier answer \(\pi rl\) to find that we can also write the surface area as \[\pi rl=\frac{\pi rh}{\cos\theta}=\frac{\pi r^2}{\sin\theta}.\]

Using the reciprocal trigonometric functions, we could also write these as \(A=\pi rh\sec\theta=\pi r^2\cosec\theta\).

Volume of a cone

The area of the base of the cone is \(\pi r^2\). The volume of any pyramidal shape is given by \[\tfrac13\times\text{base area}\times\text{perpendicular height}\] so in this case, the volume is \(\frac13\pi r^2h\). This result can be proved using calculus.