RE: Question: Proving the volume of a pyramid
March 4, 2022 at 8:51 pm
(This post was last modified: March 4, 2022 at 8:52 pm by polymath257.)
(March 3, 2022 at 2:14 pm)GrandizerII Wrote: Is there a way to prove this algebraically without the use of calculus? And without resorting to 3d visual demos, which have caused me more confusion than clarity.
Imagine a cube. Consider the top left, front vertex. Opposite that is the bottom, right, back vertex. There is a *three* fold rotation around the line joining those opposite vertices.
If you look at the pyramid consisting of the top, left, front vertex and the bottom square, that three fold rotation will take that pyramid to two other copies that, together, fill the cube.
Hence, the volume of that pyramid is 1/3 the volume of the cube.
Now use homogeneity: expand any dimension of the cube by a factor and all volumes expand by the same factor. Do the three dimensions independently, and you get the volume of a right pyramid with a square base has the volume (1/3)abc, where a,b, and c are the different lengths.
Showing that the pyramid doesn't have to be a right pyramid and the same formula applies, uses the fact that the cross sectional areas are the same, so the volumes will be as well.
In general, if you have a 'pyramid' with a base of area A and a height of h, the volume will be (1/3)Ah, so 1/3 the volume of the corresponding cylinder.
(March 3, 2022 at 8:50 pm)Jehanne Wrote:(March 3, 2022 at 2:14 pm)GrandizerII Wrote: Is there a way to prove this algebraically without the use of calculus? And without resorting to 3d visual demos, which have caused me more confusion than clarity.
I'm guessing, but, no; you need calculus, simply because horizontal area varies as a function of height. Proving that an algebraic proof does not exist may be difficult.
Nah. It was known LONG before calculus was invented/discovered. I think it was known even before Archimedes gave the best prelude to calculus the ancients had.
