mathematical and computing sciences question #3328

Kelvin C, a 100 year old male from GTA asks on March 14, 2006,


Why is it that when calculating the volume of a pyramid, such as a cone, you always have to divide by 3. This means that a cone is 1/3 of a cylinder, but why 3, why can't it be some random number. The sphere behaves the same way, but in this case, what would it be without dividing the 3 for the volume of a sphere?

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the answer

Aaron Abrams answered on March 19, 2006, A:

This is only a semi-answer to the math question in the sense that it begs another question, but the semi-answer is that the 3 you're dividing by is the dimension.

Let's look at the cone first. One way to think about its volume is to think about making the cone by starting with a region in the plane (a disk) and a point not on the plane (the cone point), and joining all the points of the plane region to the cone point. If you know how to do integrals, you should be able to show that it actually doesn't matter what shape the plane region is or exactly where the cone point is - all that matters is the area A of the plane region and the "height" h of the cone point, meaning its perpendicular distance to the plane. The volume of the cone-like thing you get will be Ah/3. If you do the same thing one dimension lower, you get Ah/2: here A stands for a length, i.e. a 1-dimensional area, and this is the formula for the area of a triangle, half the base times the height. (And note that a triangle is a 2-dimensional cone.) If you do the same thing one dimension higher you get Ah/4 (where A stands for a 3-dimensional volume, and Ah/4 is the 4-dimensional volume of the 4-dimensional cone you've constructed). And so on. If you actually do the exercise and derive the formula, you'll see how the denominator arises as the dimension.

This is a perfectly good analytic answer to the question, but the question it begs, in my mind at least, is: what's a geometric explanation for this phenomenon? In two dimensions, you can easily see why the area of a triangle is what it is: at least for acute triangles, there's an obvious rectangle of area (base)(height) whose area is clearly twice the area of the triangle. (It's not much harder for non-acute triangles.) But in three dimensions, it isn't geometrically obvious (at least to me) why the volume of the cone should be exactly a third of the volume of the corresponding cylinder.

With a ball, the situation is similar but a little trickier. You can think of making a ball by starting with its boundary sphere (just the surface) and thinking of the center point as a "cone point," and forming the ball by joining every point of the boundary sphere to the center. In computing the volume, you integrate the surface area, and a similar thing happens with the dimension. Again, you should try this (in a variety of dimensions, starting with the smallest one that makes sense) to see where the number comes from.

One thing that makes the ball trickier than the cone is that the construction is bound by a geometric constraint: for the cone you could use any initial shape but for the ball you probably want to stick with a sphere. If you're ambitious you could try a general convex shape. Another thing that makes the ball different from the cone is that usually with a ball you don't compute the volume this way, because you'd have to already know the surface area. Usually you compute it a different way (via another not particularly difficult integral) and then use the answer to get the formula for surface area. So this approach is kind of backwards from that point of view.

Anyway in both cases the question remains, if you're in dimension n, is there a geometric argument showing that some natural volume is exactly n times the volume you've constructed? That would be an elegant way to explain the n in those denominators.

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