# How to calculate the volume of an arbitrary pyramid without calculus?

I've been reading about the intuition behind calculating the volume of a pyramid by dividing the unit cube into 6 equal pyramids with lines from the center of the cube and it makes sense since all pyramids are the exact copies of each other and I'm curious how this intuition expands to rectangular prisms. Once we know the formula to calculate the volume of a pyramid we can actually see that lines from the center of the prism indeed divides the shape into 6 pyramids with equal volumes, but without knowing the formula is it possible to somehow say that?

Same question goes for other pyramids with unequal side lengths. How can you say that pyramids with same base and height have equal volumes without knowing the formula?

I'm specifically asking for a primitive method without the use of calculus or other advanced methods because I've been curious about whether or not Egyptians had a way to show it or they just got lucky, or maybe they were only interested with pyramids cut from the unit cube and not the others?

• Maybe what you are looking for is the Cavalieri Principle, though the idea, in the Mediterranean world, goes back to Hellenic times. – André Nicolas Jan 31 '16 at 18:52
• Is this What you are referencing?? For an alternate proof see here – Brevan Ellefsen Jan 31 '16 at 18:52
• @AndréNicolas it certainly looks interesting, I guess the rest is the job for historians. – none Jan 31 '16 at 19:09
• @BrevanEllefsen I have read it from somewhere else but yes it is what I'm referring to. I guess the second video is a method of exhaustion. – none Jan 31 '16 at 19:11

The idea of dividing a cube into 6 Pyramids is excellent because they are obviously equal. If you place a plane parallel to two of the sides that also cuts through the center you have created two new boxes inscribing both pyramids on either side of the plane. These pyramids were 1/6 of the volume which is to say they are 1/6 of twice the half-volume because you just cut the cube in half and 1/6 of 2 is 1/3. Now observe that the new volumes have heights unequal to the other two sides which shows that 1/3 works even if the height is different than the width of the base. Great idea for middle school without calculus and without raw memorizing of formulas derived from calculus.

The problem is essentially the motivation behind Hilbert 3rd Problem, which was solved by Dehn. The bottomline is: there is no way to use clever "cutting and putting together" pieces to find the volume of an arbitrary pyramid. You can look up "THE BOOK" by Aignier and Ziegler, there's a chapter on the Dehn invariant.

Now there are pyramids whose volume can be determined by cutting and taking multiple copies of pieces. I listed a few of those in this post

Which Pyramids have a volume which is computable by dissection?

On the other hand I don't think this is a complete list. But it is sufficient to get an idea that the formula should be right.

Lastly, the best proof which uses "the method of exhaustion" (which is a precursor to calculus; you basically need limits but not integration) is that of Euclid. See
https://mathcs.clarku.edu/~djoyce/elements/bookXII/bookXII.html The volume of pyramids is discussed in Proposition 3 to 5.

I wanted to (roughly) calculate how much wood I had cut in a reasonably pyramid like pile. When looking up the formula, I too was struck by the simplicity of the bwh/3 answer, and wanted to understand the proof.

The use in the proof of the cube sliced internally to create 6 equal pyramids is a classic bit of Eureka, I would suggest, pure genius. After that, you can confirm to yourself that the pyramids are indeed equal by simple Pythagoras.

Finally, I wanted to generate a simple supporting proof to get it straight in my mind and came up with this (c=cube, p=pyramid):

b x w x hc= Vc =6Vp

b x w x 2hp= Vc = 6Vp

b x w x 2hp/6 =Vp = b x w x hp/3

The second part of your question, about rectangular bases, would seem to be an acceptable intuitionistic step from the above, at least to my mind. I must admit that the fact that, if you keep the apex at the same height and skew the pyramid around so that you have unequal triangular sides, the formula for volume still works (using perpendicular height always), was less intuitive for me.