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

- Parallel cross-sections of a cylinder have constant area. Therefore its volume is , where is the area of the base.
- Parallel cross-sections of a cone are scaled in both directions by the -coordinate. This multiplies the sectional area by and therefore the volume is .
- There is something in between: we can scale cross-sections in
**one direction only**. Then the sectional area is multiplied by and the volume is .

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 and in , for define

Each is convex. If the sets and are cone and cylinder of equal height and with the same convex base , then all 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 is always a **cubic polynomial** in , even if the convex sets are defined by some transcendental functions. I still do not find this fact intuitive.

Let’s check it on the example of and being a cone and a cylinder of height and radius . Obviously, every 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 :

hence

It so happens that 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 , the volume of is equal to , same as the volume of James Tanton’s solid. Since is pretty small, the shape looks much like a cone.

When , the volume of is equal to , which is exactly halfway between the volumes of cone and cylinder. The solid exhibits harmonious balance of both shapes.