Spherical Surfaces and Hat Boxes

To round off our series on round objects (see the first and second posts), let’s compute the sphere’s surface area. We can compute this in the same way we related the area and circumference of a circle two weeks ago. Approximate the surface of the sphere with lots of small triangles, and connect these to the center of the sphere to create lots of triangular pyramids. Each pyramid has volume \(\frac{1}{3}(\text{area of base})(\text{height})\), where the heights are all nearly \(r\) and the base areas add to approximately the surface area. By using more and smaller triangles these approximations get better and better, so the volume of the sphere is $$\frac{4}{3}\pi r^3 = \frac{1}{3}(\text{surface area})\cdot r,$$ meaning the surface area is \(4\pi r^2\). (This and previous arguments can be made precise with the modern language of integral calculus.)

Dividing a sphere into many pyramids
Dividing a sphere into many pyramids connected to the center allows us to relate the sphere's surface area and volume.

Here’s an elegant way to rephrase this result: The surface area of a sphere is equal to the area of the curved portion of a cylinder that exactly encloses the sphere. In fact, something very surprising happens here!:

Archimedes’ Hat-Box Theorem: If we draw any two horizontal planes as shown below, then the portions of the sphere and the cylinder between the two planes have the same surface area.

Archimedes' Hat-Box Theorem
Any two horizontal planes cut off a band on the sphere and another band on the enclosing cylinder. Archimedes' Hat-Box Theorem says that these bands have the same area. (The planes shown here have heights \(0.4\cdot r\) and \(0.6\cdot r\) above the equator.)

We can prove this with (all!) the methods in the last few posts; here’s a quick sketch. To compute the area of the “spherical band” (usually called a spherical zone), first consider the solid spherical sector formed by joining the spherical zone to the center:

Spherical sector
The Hat-Box theorem can be proved by relating the area of the spherical zone to the volume of this spherical sector.

By dividing this into lots of triangular pyramids as we did with the sphere above, we can compute the area of the spherical zone by instead computing the sector’s volume. This volume can be computed by breaking it into three parts: two cones and the spherical segment between the two planes (on the left of the next figure). Compute the volume of the spherical segment by comparing (via Cavalieri’s Principle) to the corresponding part of the vase (from the previous post), which can be expressed with just cylinders and cones.

Volume of a spherical segment via Cavalieri's Principle
Computing the volume of a spherical segment via Cavalieri's Principle

See if you can fill in the details!

2 thoughts on “Spherical Surfaces and Hat Boxes

  1. This and your previous post were very helpful, thanks, and nice illustrations! This has allowed me to prove volume and surface area formulas for spherical cap WITHOUT calculus, an interesting exercise. Cavalieri’s Principle proves the volume using the ‘Hat Box’ and Vase constructions, similarly to how it proves volume of sphere. To establish surface area from the volume, it works along same lines as it does for the sphere, by constructing a polyhedral approximation using latitudinal and longitudinal subdivision into pyramidal elements which are tangent planes to radial vectors located at the center of the elements (4 adjacent element faces meet at a point, so forming a polyhedral surface). The polyhedron is then cut off by the cone projection. These pyramidal elements (as conic solids) have volume EXACTLY (1/3)*base area*r, where r = radius of sphere. Adding their volumes approximates volume of spherical cone (ie cone + cap), and adding their base areas approximates the surface area of cap. The relationship V = (1/3)*r*A holds all the way through in the approx’s as n -> infinity. Because of the geometry, the nth vol approx Vn is arbitrarily close to true vol V, but any subsequence of (Vn) is ‘AC’ to V also (these approx’s become perfectly ‘smooth’ – ie angles between adjacent faces -> 180 degrees). Hence formally lim Vn = V. Likewise lim An = A. Hence for the spherical cone we obtain V = (1/3)*r*A, where A is the cap surface area. From that and the known cap and cone volumes we obtain A. Archimedes’ Hat-Box Theorem then follows easily as a corollary. But I didn’t see how I could use Cavalieri to get the volume of cap to begin with, your post helped with that!

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