4 ms·

If a triangle is 1/2 a rectangle, a cone is 1/3 a, cylinder, is there some kind of n-d hyperpyramid that is 1/n of its bounding hypercube? As in, does the 1/3 h

• Chinjut 3y ago
  Yes, that's exactly what happens. Think of a point in an n-dimensional hypercube and the pyramids made by connecting it to the n many opposite faces. These pyramids each have volume 1/n times the hypercube's volume. And in the same way, in n dimensions, any tapering figure has volume given by its height times its base divided by n.
• tylerneylon 3y ago
  Yes; one way to see this is with a little calculus. It all boils down to the fact that the indefinite integral of x^(n-1) is x^n / n. (That's where 1/n comes in.) Suppose we have an n-dimensional pyramid. It has an (n-1)-dim'l base with volume B, and that base tapers to a point; for simplicity, it has height 1. Take cross-sections as we travel from the base to the height. How big is a cross section at height y? Each cross section is a (n-1)-dim'l shape, and its volume is B * y^(n-1) because this is how volumes scale in dimension n-1. [Examples: In 2D if you x2 the sides of a shape, it's "volume" (area) is x4. In 3D if you x2 the sides of a shape, it's volume is x8; in 4D it would be x16, etc.] Now take the integral to add all of these cross-sections: integral(from 0 to 1 of B * y^(n-1)) = B * y^n / n evaluated from 0 to 1 = B / n. If the shape is scaled along the height dimension by h, then we get the more general formula: volume = h * B / n.