Volume of a Torus

I once received an email message from someone who wanted to know the volume of a torus. An interesting question, thought I. Actually, I had already been told the answer, but I had never bothered to investigate it and prove it. If I were to do this again, the easiest way probably would be to use Pappus’ Theorem for the volume of a solid of revolution. If you have not heard of it, look it up in a calculus text. It is not terribly complex, but it requires a good understanding of integral calculus.

As it happened, I did not use Pappus’ Theorem. Instead, I started by using a much less efficient method. In doing so, I stumbled onto a gem of an explanation. It is similar to a technique used by Archimedes in the third century BC. He would often deduce geometric properties of a figure by comparing it with a simpler, better known figure. He would refer to planar regions, and even line segments, as thought they had mass and weight. Archimedes deduced the volume of a sphere by comparing it with a cylinder and a cone. For an explanation of this, see the article “The Method of Archimedes” by John Del Grande, Mathematics Teacher 86(3), March 1993.

As it happened, I wrote the first version of this in 1997, while at the same time getting to know a nice graphics software package called Fields and Operators, published by Lascaux Graphics. I rather enjoyed it, and that is how I worked up most of the images here. New computers and operating systems often are incompatible with old software, and this is one of many losses for me. I just added an extension to this article, and you will notice an abrupt change in the images.

A torus is a solid of revolution. It is formed by rotating a circle about a line that is in the plane of the circle, but not intersecting the circle. A torus has the shape of a doughnut. Let r be the radius of the circle being rotated, and let R be the distance from the center of the circle to the axis of rotation.

Orient the torus so that its widest cross-section is aligned with the x‑y plane and its axis is the z‑axis. Now intersect the torus with a plane parallel to the x‑y plane. What is the region of intersection?

The intersection is a washer-shaped region called an annulus. It is the region between two concentric circles, so its area is the difference between the circle areas. Given the z‑coordinate of the intersecting plane, we should be able to compute the area of the annulus. If we look at a profile view, we can see some simple relationships.

With dimensions shown above, the area of the annulus is simply the difference of areas of its two circular boundaries.

Now comes the comparison part. Here is a cylinder with radius r, and length 2pR. The cylinder’s axis lies in the x‑y plane. See what happens when it is cut by the same plane that intersected the torus.

The plane’s intersection with the cylinder is a rectangle. Its length is 2pR, the same as the length of the cylinder. Notice that a vertical cross-section of the cylinder matches a corresponding cross-section the torus, because they have the same radius, r. So the width of the rectangle is 2a.

This is the same as the area of the annulus. For any z, a horizontal plane intersects equal areas with the torus and the cylinder, although their shapes are not at all similar. It can be imagined that arbitrarily thin slices are being shaved off of the torus and the cylinder, and are made of the same material. Each time the cutting plane passes through the solids, it cuts slices which are equal in area and thickness. If we pile the slices on oposite pans of balance scales, they must remain balanced. The objects have the same weight, hence the same volume. We already have a simple formula for the volume of a cylinder.

This is hardly a rigorous proof, but I am hoping that it conveys a qualitative understanding. The notion of cutting objects into thin, measurable slices is essentially what integral calculus does. Archimedes was practicing this method about 1900 years before the era of Leibnitz and Newton.

Extension

Begin with a sphere of radius r. Let a cylindrical hole be drilled through it. The hole has a vertical axis of length L, and the axis of the hole goes through the center of the sphere. What is the volume of the solid that is left behind? I say it is equal to the volume of a sphere of diameter L. Prove me right or prove me wrong.

Here they are in profile. As with the torus problem, let the two solids be cut by the same horizontal plane. Their cross-sections are an annulus and a circle. Show that the cross-section areas are equal.

If I am right – and I am – there is a certain counterintuitive result. The radius of the sphere does not even matter. Here are three spheres, all with drill holes of equal length. The same volume is left behind in all three cases.

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