Question: Proving the volume of a pyramid

[Jehanne]

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.

[brewer]

I typically look at the nob to see it it's set at 11.

[Fireball]

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.

The link provides a straightforward explanation of the calculation. A proof thereof would simply be some statements justifying why the measures are true.

[polymath257]

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.

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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.

[The Valkyrie]

I typically look at the nob to see it it's set at 11.

What you do in the privacy of your own room in no-one's business but your own.

[Rev. Rye]

You need a pyramid and a tape measure.

And as Dr Ben Carlson will tell you,  a lot of grain to fill the pyramid with so as to know its volume.

... I wonder if storing a mummy in a grain silo changes the taste. Maybe ancient Egyptian bread was like Mezcal.

[Ranjr]

The first question I have for this measure, is, why are these super powered people visiting the planet only to stack stones?  Those are the margins that concern me.

[GrandizerII]

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.

Thanks, polymath. But I think the issue with me is that I can't visualize that three-fold rotation well. Even if I were to watch a video/animation of that occurring, I still would probably struggle to have that intuition once I look away from the screen. Thanks for coming up with something simpler than other stuff I've checked online, though.

I'll get to Fire's article soon. If still struggling, maybe it's easier for me to just go with the calculus route then.

[polymath257]

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.

Thanks, polymath. But I think the issue with me is that I can't visualize that three-fold rotation well. Even if I were to watch a video/animation of that occurring, I still would probably struggle to have that intuition once I look away from the screen. Thanks for coming up with something simpler than other stuff I've checked online, though.

I'll get to Fire's article soon. If still struggling, maybe it's easier for me to just go with the calculus route then.

Imagine holding the cube between your thumb and forefinger by opposite corners of the cube. if you have a pair of dice, take one and do it. You will find that you can rotate around the axis between your fingers and the resulting rotation is three fold.

There are three edges coming out of each vertex. The three fold rotation  cycles those edges out of the opposite corners. it also cycles the faces that come together at those corners.

It *is* a difficult rotation to imagine. But it is well worth it.

If you take a (planar) cross section perpendicular to the line between opposite corners and half way between them, the cross section on the cube is a hexagon (!).

[Fireball]

Yup. If you draw a line segment from alternating vertices of a hexagon to the center, it will look like a cube with a vertex facing you.