Interpolating between a cone and a cylinder

This post is inspired by James Tanton’s video which is summarized in bulleted items below. For convenience normalize all solids to height {1}.

  • Parallel cross-sections of a cylinder have constant area. Therefore its volume is {\int_0^1 A z^0\,dz =Ah}, where {A} is the area of the base.
  • Parallel cross-sections of a cone are scaled in both directions by the {z}-coordinate. This multiplies the sectional area by {z^2} and therefore the volume is {\int_0^1 A z^2\,dz =\frac{1}{3}Ah}.
  • There is something in between: we can scale cross-sections in one direction only. Then the sectional area is multiplied by {z} and the volume is {\int_0^1 A z\,dz =\frac{1}{2}Ah}.

Taking the base to be a circle, we get this shape:

Pinched in one direction
Pinched in one direction

It looks… odd. Not even convex:

Convexity is lost
Convexity is lost

But cones and cylinders are convex whenever the base is a convex region. Can we interpolate between them within the class of convex sets?

The answer is: yes, if we use the Minkowski sum. Given two convex bodies {A_0} and {A_1} in {{\mathbb R}^3}, for {0< t<1} define

\displaystyle  A_t =(1-t)A_0+tA_1 = \{(1-t)a_0+ta_1 : a_0\in A_0, a_1\in A_1\}

Each {A_t} is convex. If the sets {A_0} and {A_1} are cone and cylinder of equal height and with the same convex base {B}, then all {A_t} have the same base and the same height.

Finding the volume of a convex body in Calculus II/III can be a tiresome exercise with trigonometric substitutions and whatnot. It may come as a surprise that the volume of {A_t} is always a cubic polynomial in {t}, even if the convex sets {A_0, A_1} are defined by some transcendental functions. I still do not find this fact intuitive.

Let’s check it on the example of {A_0} and {A_1} being a cone and a cylinder of height {1} and radius {r}. Obviously, every {A_t} is a solid of revolution. Its profile is the Minkowski sum of appropriately scaled triangle and rectangle. This is how I imagine it:

  • Mark a point on the rectangle, for example the midpoint of the bottom edge.
  • Move the rectangle around so that the marked point stays within the triangle.
  • The region swept is the Minkowski sum.
Minkowski sum
Minkowski sum

The volume can now be computed directly. I used the venerable method of cylindrical shells to find the volume of {A_1\setminus A_t}:

\displaystyle 2\pi \int_{tr}^t x(x/r-t)\,dx = \frac{\pi r^2}{3}(2-3t+t^3)

hence

\displaystyle  \mathrm{Vol}\, (A_t) = \frac{\pi r^2}{3}(1+3t-t^3)

It so happens that {\mathrm{Vol}\,(A_t)} is a concave function in this example, although in general only its cubic root is guaranteed to be concave, by the Brunn-Minkowski inequality.

When {t\approx 0.168}, the volume of {A_t} is equal to {\pi r^2/2}, same as the volume of James Tanton’s solid. Since {t} is pretty small, the shape looks much like a cone.

1/2 of area*height
Half of Area*Height

When {t\approx 0.347}, the volume of {A_t} is equal to {2\pi r^2/3}, which is exactly halfway between the volumes of cone and cylinder. The solid exhibits harmonious balance of both shapes.

2/3 of area*height
2/3 of area*height

1 thought on “Interpolating between a cone and a cylinder”

  1. How do I get bodies from a fixed base but whose volume 1/4, 1/5, etc, the volume of the cylinder?
    Something like k*Volume cylinder where 0<=k<=1

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