Gabriel’s Horn

(Edited 6/7/2020 for improved typesetting)

Here is one of my favorite “paradoxes” in mathematics. Many students learn it in a first year calculus course. It is called Gabriel’s Horn.

First, take the function $f(x)=\dfrac 1x$ on the domain $[1,\infty)$ . So, a graph of this function begins at $(1,1)$ then quickly slopes down towards the $x$ -axis, where it approaches horizontal. A plot is shown below.
prerotation
Imagine taking the curve and rotating it around the $x$ -axis, forming an infinitely long shape that looks like the bell of a trumpet, or horn. After doing so, we get a shape that looks roughly like this:
geogebra-export(1)
There are two things we wish to determine about this new three-dimensional figure: its volume, and its surface area.

We can think of the volume as how much stuff is needed to fill it, and the surface area as how much paint we need to coat the surface. We will show this figure has finite volume but infinite surface area. In other words, you would need an infinite amount of paint just to cover the inside surface with a layer of paint. However, if we really wanted to paint the inside, we could take our finite amount of paint and just pour it into the horn. This would fill up the infinite horn with a finite amount of paint, thus painting the inside.

We do this calculation in the standard calculus way: Let’s consider an arbitrarily thin vertical slice of the horn at the point $x_1.$ The disk we get as a result is circular with radius $r= \dfrac{1}{x_1}.$
finalHorn
So, for any vertical slice taken from our horn we can find its radius, and thus both its circumference $(2\pi r)$ and area $(\pi r^2).$

We integrate over the entirety of the horn: Integrating using circumference gives us surface area, while integrating using area gives us the volume. For the sake of completeness, I’ll do the limits on the improper integrals that result.

Substituting $r = \dfrac 1x$ , our surface area integral computation goes as follows.

\begin{aligned} \text{Surface Area} &= 2\pi \int_1^\infty \frac{\mathrm dx}{x}\\ &= 2\pi \lim\limits_{b\to\infty}\int_1^b \frac{\mathrm dx}{x}\\ &= 2\pi \lim\limits_{b\to\infty} (\ln b-\ln 1)\\& = \infty \end{aligned}

Note that since $\ln 1=0$ we just get the term $\ln b$ going off to infinity. As promised, the surface area is infinite. We do a similar computation in volume.

\begin{aligned} \text{Volume} &= \pi \int_1^\infty \dfrac{\mathrm dx}{x^2} \\ &= \pi\lim\limits_{b\to\infty} \int_1^b \frac{\mathrm dx}{x^2}\\ &= \pi\lim\limits_{b\to\infty}\left. -\frac1x \right|_1^b \\ &= \pi\lim\limits_{b\to\infty}\left(-\frac 1b +1\right)\\ &= \pi \end{aligned}

Here, the key is that our $b$ term is in the denominator, taking that term to $0$ overall, giving us finite volume.

There are many cool problems like this, some with significantly less background necessary to formulate. These will be fun to mix in with other types of posts.

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