The Volume of a Cone

A few days ago, we pulled on a historical thread and wound up browsing the early volumes of The Mathematical Gazette. Doing so, we stumbled across a “mathematical note” from 1896 by Alfred Lodge, the first president of the Mathematical Association. Lodge’s note provides a simple derivation for the volume of a cone. Such arguments don’t vary all that much but, however we missed it, we’d never seen the derivation in the very elegant form presented by Lodge. Here is Lodge’s argument, slightly reworded.

Imagine a cone of height H, and imagine slicing the cone at some level, to give a new cone of height h. Whatever h, any such cone is similar, and so the volumes will scale as \boldsymbol{h^3}. That is, for all such cones we have

    \[\boldsymbol{\color{blue}V(h) = kh^3\,,}\]

and our problem is to find \boldsymbol{k}, the constant of proportionality. To do so, think of \boldsymbol{h} close to \boldsymbol{H}. Then the difference in the cones is approximately a cylinder of height \boldsymbol{H - h}, and with the same base as that of our original cone. So, calculating the volumes,

    \[\boldsymbol{\color{blue}\mbox{\bf Base}\times(H-h) \, \approx \,V(H) - V(h) \, = \, kH^3 - kh^3  \,.}\]

Dividing by \boldsymbol{H - h} and factoring the difference of two cubes,  we then have

    \[\boldsymbol{\color{blue}\mbox{\bf Base} \, \approx \, k\frac{H^3 - h^3}{H-h}\, = \,  k\frac{(H-h)\left(H^2+Hh+ h^2\right)}{H-h} \, = \,  k\left(H^2+Hh+ h^2\right)\,.}\]

Letting \boldsymbol{h \to H}, we find

    \[\boldsymbol{\color{blue}k \, = \,\frac{\mbox{\bf Base}}{3H^2}  \,.}\]

Plugging back in and setting \boldsymbol{h = H} gives the desired formula:

    \[\boldsymbol{\color{blue}V(H) = \frac13 \times \mbox{\bf Base} \times H\,.}\]

The reader is invited to ponder other shapes and other dimensions.

13 Replies to “The Volume of a Cone”

  1. What do you think of the approach taken by Lockhart in Measurement?
    A square pyramid with h = 1/2 sqrt B has volume 1/3 Bh by symmetry as 6 of them make up a cube of twice the height.
    Any other height square pyramid will scale linearly with its height based on the change in area of the triangular cross sections.
    A circular cone with the same base area will have the same area at any horizontal cross section as the square pyramid again area formulae and similar triangles.
    As the horizontal cross section at any height is equal the volume must be equal.

    1. Thanks, Stan. I’ll take a look and comment more. In brief, the individual ingredients are nice (and standard), but in sum it feels too busy. I can see a good Year 8 or Year 9 student getting a solid sense of Lodge’s argument, but I can’t see them getting as much from what you’ve outlined.

  2. I mean, that \simeq is doing a bit of work here, I wouldn’t call that a rigorous argument. I think making it rigorous would be a nice thing to do for students at the right level (not high school).

    These kinds of arguments are why you get things like the troll math meme.

  3. You once wrote something like, it takes a special kind of idiot to use D and d in the same formula. Is H and h any different?

    1. How dare you suggest that I may be inconsistent! Whose blog do you think this is!

      OK, having calmed down from that horrible effrontery …

      Thanks, Ben 10. You’re referring to my comment on this post. The difference is, my h and H and referring to different values of the same height variable, whereas in the PISA question D and d are totally distinct things, a drip rate and a drop factor.

      Although, to be fair, Andreas Schleicher is both a drip and a dropkick, so maybe using both D and d is reasonable.

    1. Yes. You can think of it as showing the derivative of the area is the integrand. There’s very general formulas of this type. But the similarity makes it a little different, and quicker.

      1. Thinks: Here’s an opportunity to tangent again. (from tango, tangere, tangi, tactum)

        You know how western calculus started with completed infinities, infinitesimals and infinite series, and how after many years and much heartache, these were replaced by the limit concept. The conventional understanding is that Greek mathematics went the other way, starting with the method of exhaustion (a rigorous limit based method) and later using infinitesimals. Specifically, Archimedes in “On Method”, thought of a sphere as an infinite collection of plane discs and weighed the discs using levers to arrive at V = (4/3) \pi r^3. However, we have almost nothing of Greek maths writing before Euclid, ~300 years of undocumented effort, Thales to Eudoxus. My thesis is that Greek mathematics went through a similar orbit to our calculus, and started with infinitesimals; the history of this has been lost. Archimedes was happy to go back to the ancient way operating, just to obtain informal results, but made it clear that the method was just suggestive.

        Imagine a far future civilization, that is researching our achievements in mathematics using only our current textbooks, missing the first 300 years. They would see a plethora of calculus books that all start from the limit concept. Then the odd advanced book that expounds the Robinsohn concept of infinitesimal numbers. They too would get it arse-about.

        We will probably never know what the Greeks did in the missing period. But it is fun trying to reconstruct an earlier free-wheeling approach. For example, can you find A = \pi r^2 for a disc by adding 1-dimensional curves. No, don’t use infinitesimal triangles, plain curves will do.

        1. Yes there is no direct evidence at all that the Greeks went from infinitesimals to exhaustion. And there is no direct evidence at all that the Greeks went from exhaustion to infinitesimals. The evidence either way has not survived. So we are required to assign a priori probabilities of 50% and 50%. Now apply the fact that the experiment that we do know, the invention of calculus in Western Europe, used the first route. And apply the fact that, at least for simple examples, the first route is easier to comprehend.

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