Gabriel’s Horn (guest post)

Last week I was intrigued by this post:

Gabriel's Horn is a solid you get by rotating the hyperbola y=1/x (with x>1) about the x-axis.

Having finite volume (π) and infinite(!) surface area, it leads to the apparent paradox:

"You can fill it with paint, but you cannot coat it." pic.twitter.com/6xLB71wrEy
— Tamás Görbe (@TamasGorbe) October 19, 2022

I asked, on Facebook, whether filling it with paint would essentially be painting it on the inside and I had a suspicion that my good friend Art Guetter (Professor of Mathematics at my institution) would help me learn something. I was right, and he’s been kind enough to type up his thoughts for this post. Here he shows that it’s pretty tough to use a finite amount of paint to paint anything infinitely long.

Guest post from Art Guetter

Gabriel’s horn is a solid created by rotating the graph of $f(x) = 1/x$ , defined on the interval $(1,\infty)$ , around the $x$ -axis. Two “paradoxes” arise. In the first, the horn has a finite volume, despite being created by rotating a region with infinite area around an axis. The second is that the horn has finite volume but infinite surface area, leading to the apparent paradox that the horn could be filled with a finite volume of paint, yet the paint would not be sufficient to coat (that is, paint) the surface. The resolution of this “painter’s paradox” is that the thickness of the paint would need to decrease to 0 in the limit as $x$ tends to infinity. The assumption being made here is that painting requires a uniform thickness of paint. Note that I can paint the entire plane if I am allowed to decrease the thickness of paint as I move far from the origin.

So could I paint an infinite solid of revolution (to a uniform depth $h$ ) if the surface area were finite? As a first example, replace $f(x) = 1/x$ from Gabriel’s horn with a piecewise constant function $f(x) = r_n$ when $n < x < n+1$ for $n = 1,2,3,\ldots$ , and the constants $r_n$ to be determined later. The surface will consist of an infinite collection of right circular cylinders, and each cylinder will have surface area $2 \pi r_n$ . If the $r_n$ are chosen so that the sum $\Sigma r_n < \infty$ , can I paint the surface with a finite amount of paint? The answer appears to be “yes”, but this involves the assumption that I roll each cylinder open, so that the amount of paint used is simply the surface area multiplied by the thickness of the paint, say $h$ . (Each cylinder can be rolled open without issue because they have thickness 0.)

What about painting if the cylinders aren’t rolled out? I will assume that painting to a thickness $h$ means that the depth of paint at any point is $h$ measured along the normal, in the outward direction. The amount of paint needed to paint one of the cylinders is then given by $\pi [(r_n + h)^2 - r_n^2] = \pi [2 r_n h + h^2] = 2 \pi r_n [h + h^2/(2r_n)] = A_n [h + H h^2]$ , where $A_n$ is the surface area of the cylinder and $H = 1/(2r_n)$ is the mean curvature of the cylinder. Summing this over $n$ will lead to an infinite volume of paint, no matter how fast the $r_n$ tend to 0.

A more general theorem has that the volume of a surface that has been thickened by an amount $h$ in the direction of the normal to the surface (assuming that $h$ is small enough that there is no self-intersection) is given by

$V = h \cdot SA + h^2 \cdot \int H \; dA +\frac{1}{3} h^3 \cdot \int K \; dA$

where $SA$ is the total surface area, $dA$ is the surface element, $H$ is the mean curvature of the surface, and $K$ is the Gaussian curvature. (These are constant for the cylinder, with values $H = 1/(2 r_n)$ and $K = 0$ .) The amount of paint needed to paint a surface to uniform thickness depends on the curvature of the surface.

For a surface created by revolving the curve $y = f(x)$ around the $x$ -axis, the values of $H$ and $K$ depend only on $x$ and are given by (in general and then for $f(x) = 1/x$ )

$H = \frac{1 + f'(x) - f(x)f''(x)}{2 f(x) (1 + f'(x)^2)^{3/2}} = \frac{x^7 - x^3}{2(1 + x^4)^{3/2}}$

$K = -\frac{f''(x)}{f(x) (1 + f'(x)^2)^2} = -\frac{2 x^6}{(1 + x^4)^2}$