There are a number of ways to derive the volume
(V) of a cone or pyramid with a height h, and a base with area
B. Here is how I did it, in high school. Return to my Mathematics pages
Go to my home page
Note: If your WWW browser cannot display special symbols, like ² or 2 or ±, then click here for the alternative Cones and Pyramids page.
© Copyright 1998, Jim Loy
There are a number of ways to derive the volume
(V) of a cone or pyramid with a height h, and a base with area
B. Here is how I did it, in high school.
I divide the object (cone or pyramid) up into many layers (n of them, actually). And I approximate each layer with prism-like object, the same shape as the base, with the sides vertical, not slanting. By "prism-like," I mean that if the original object is a square pyramid, then each layer is now a rectangular solid. If the original object is a circular cone, then each layer is now a circular cylinder. If the original object is some pyramid, with an odd-shaped base, then each layer is now a prism, with a base of the same odd shape.
Let's say that each little layer has a height of x. Then, h (the height) is equal to nx. In other words, x=h/n. Each part (side, diameter, whatever) of the base of each layer follows an arithmetic progression, from the smallest at the top, to the largest at the bottom. As i (a variable that we are using to count the layers) varies from 1 to n, the area of the base of each layer is Bi²/n². And the volume of layer i is Bhi²/n³. The sum of all of the layers is the sum (as i goes from 1 to n) of this Bhi²/n³. Well, for any given cone or pyramid, all of those letters are constants, except for i. So, that sum is Bh/n³ times the sum of all of i² (where i varies from 1 to n).
It turns out that it is fairly easy to show that this sum of i², where i varies from 1 to n, is n(n+1)(2n+1)/6. I will not prove that here. But, it can be proven by mathematical induction, as well as other ways.
So, the volume of our approximation of a cone, or pyramid, is Bhn(n+1)(2n+1)/6n³. Dividing the numerator by n, three times, we can rewrite that as Bh(1+1/n)(2+1/n)/6.
We want our approximation of a cone, or a pyramid, to no longer be an approximation, but to be exact. To do that, we need n to become infinite. In that case 1/n becomes zero. And the above expression simplifies to V=Bh(1)(2)/6=Bh/3. The volume of a cone or pyramid is 1/3 the area of the base times the height.
Note: The cone or pyramid does not have to be a right cone or pyramid. In other words, the cone or pyramid can lean. The above arguments are not affected by that. Also, a cone can have an ellipse, or some other curve, for a base. And a pyramid can have any polygon, irregular and/or non-convex, for a base.
See Cone-Shaped Milkshake Glasses.