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.
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 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?
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:
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.
