Nontrivial trivial integrals

Question

Consider two propositions in geometry:

• Circumscribe a right circular cylinder about a sphere. The surface area of the cylinder between any two planes orthogonal to the cylinder's axis equals the surface area of the sphere between those two planes. (Archimedes showed this.)
• Let a curve running from the south pole to the north pole on a perfectly spherical earth meet every meridian of longitude at the same angle $\gamma$. Then the length of the curve is proportional to $\sec\gamma$. (In particular, it's infinite if $\gamma$ is a right angle, and one could even go beyond a right angle and consider it an oriented length, so that the oriented length is negative if the curve goes from the north pole to the south pole. In that case, the infinite length would be the $\infty$ that's at both ends of the real line rather than $+\infty$ or $-\infty$, so the length depends continuously on $\gamma$.)

Both propositions admit the same kind of proof: For the first, consider what happens when the distance between the two planes is infinitesimal; then show that if it works then, then it also works for larger-than-infinitesimal distances. For the second, show that the length of an infinitesimal increment of the curve is $\sec\gamma$ times the latitude-component of the distance, just by drawing an infinitesimal right triangle.

You might think that in both cases one is finding an integral, but it's an integral of a constant: just the constant itself times the length of the interval over which one integrates. So it's as if you don't need to know about integrals and you certainly don't need to find any antiderivatives, but the method of proof doesn't work for larger-than-infinitesimal quantities.

One would like to say things like "energy is force times distance; it is only when the force is not constant that one must use integrals and multiply the force at an instant by the infinitesimal distance and then integrate". But if proving the force is constant itself requires an infinitesimal viewpoint, then despite the triviality of computing the integral by multiplying two numbers, it seems necessary to the logic of the argument to view it as an integral.

So:

• What other instances besides these two are known? (I suspect there are many in geometry.)
• What else of interest can be said about this phenomenon?

Answer 1

Here's a problem that I think has a similar flavor:

Suppose that a bounded convex solid $S$ in $\mathbb{R}^3$ is positioned in a random orientation. Show that the expected area of its "shadow" onto the $xy$-plane is $1/4$ the surface area of $S$.

To solve this, we invert the problem: suppose that our set $S$ is fixed, and the angle of the light (and projected plane) are chosen at random.

Now, for each infinitesimal piece of boundary of $S$, say that its contribution to the shadow is half its projection onto the plane. Then, because there is a "near" and "far" portion of the convex set, the sum of all contributions across the surface of $S$ will be the area of its shadow, since we double-cover the shadow and divide by $2$. (This works because $S$ is convex, so every part of $S$ is either directly in the light, or directly on the dark side - every normal vector from the interior of the shadow passes through $S$ twice.)

So we just need to integrate, for each infinitesimal bit of surface, the expected size of its shadow when light is shone on it at a random angle. But this is just proportional to the surface area of that piece! So the answer must be linear in the surface area of $S$ when we integrate, and by considering the case of a spherical $S$ we see that the constant of proportionality ends up being $1/4$.

---

As a remark, this argument carries through in any dimension (we didn't use anything unique to $\mathbb{R}^3$ here), and in the two-dimensional case it gives us a constant of proportionality of $1/\pi$. When we apply this to the case of curves of constant width, we immediately recover Barbier's theorem. In fact, this proof shows us in a sense the "correct" generalization of curves of constant width into $n$ dimensions: they are the convex sets whose projections have constant $n-1$-dimensional measure! Barbier's theorem does generalize to such sets via the above proof, but when $n>2$ this is a different notion than "constant width" and so an analogous result does not hold.

Answer 2

How about Mamikon's Sweeping-Tangent Theorem? The simplest example is the area of the annular ring between two concentric circles, where it can be shown that the area only depends on the length of the chord $a$. Thus one can imagine the inner circle shrinking to a point, and simply compute the area of an ordinary circle.