# 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:

It looks… odd. Not even convex:

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.

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.

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.