Gabriel’s horn is the surface created by rotating 1/x around the x-axis. It is often introduced in calculus classes as an example of a surface with finite volume and infinite surface area. If it were a paint can, it could not hold enough paint to paint itself!
This post will do two things:
- explain why the paradox works, and
- explain why it’s not paradoxical after all.
Rather than working out the surface area and volume exactly as one would do in a calculus class, we’ll be a little less formal but also more general.
Original function
When you set up the integral to compute the volume of the solid bounded by rotating the graph of a function f, the integrand is proportional to the square of f. So rotating the graph of 1/x gives us an integral whose integrand is proportional to 1/x² and the integral converges.
When you set up the integral to compute the surface area, the integrand is proportional to f itself, not its square. So the integrand is proportional to 1/x and diverges.
Generalization
For the volume to be finite, all we need is that f is O(1/x), i.e. eventually bounded above by some multiple of 1/x, and in fact we could get by with less.
For the area to be infinite, it is sufficient for the function to be Ω(1/x), i.e. eventually bounded below by some multiple of 1/x. An as before, we could get by with less.
So to make another example like Gabriel’s horn, we could use any function in Θ(1/x), i.e. eventually bounded above and below by some multiple of 1/x. So we could, for example, use
f(x) = (x + cos²x) / (x² + 42)
If you’re unfamiliar with the notation here, see these notes on big-O and related notation.
Resolution
Now back to the idea of filling Gabriel’s horn with paint. If we spread the paint on the outside of the can with any constant thickness, we can only cover a finite area, but the area is infinite, so we can’t paint the whole thing.
The resolution to the paradox is that we’re requiring the paint to be more realistic than the can. We’re implicitly letting the material of our can become thinner and thinner without any limit to how thin it could be. If we also let our paint spread thinner and thinner as well at the right rate, we could cover the can with a coat of paint.
OK, I’ll bite. Since the volume is finite, we can fill the can. Doesn’t this effectively “paint” the inside of the can? So we can paint the inside, but not the outside?
If the paint has any minimum thickness, it won’t reach the bottom of the can.
This could even be the case with a physical can. Once the bottom is narrow enough, paint won’t keep going.