Gabriel's Horn

Gabriel's horn is a shape with the paradoxical property that it has infinite surface area, but a finite volume.

Consider the surface area and volume of the solid formed by rotating the region bounded by the $x$-axis, $x =1$, and $y = \frac{1}{x}$, around the $x$-axis. This solid is called Gabriel's horn.

The volume of the solid of revolution can be found using the disk method:

$V = \pi \int_1 ^a \frac{dx}{x^2} = \pi \left( 1 - \frac{1}{a} \right).$

Now consider what happens as we allow $a$ to approach infinity:

$\displaystyle \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi.$

The surface area of a solid of revolution is given by the formula

$S = 2\pi \int_a^b r(x)\sqrt{1+\left( \frac{dy}{dx} \right )^2 } \, dx.$

In this case, since $\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2},$ that gives us

$\begin{aligned} S &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+\left( -\frac{1}{x^2} \right )^2 } \, dx \\ &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx. \end{aligned}$

This integral is hard to evaluate, but since in our interval $\sqrt{1+ \frac{1}{x^4}} \geq 1$ and $\frac{1}{x} > 0$,

$2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx \geq 2\pi \int_1^a \frac{1}{x} \, dx.$

It follows that:

$\begin{aligned} S &\geq 2\pi \int_1^a \frac{1}{x} \, dx \\ S &\geq 2\pi \ln a. \end{aligned}$

But $\displaystyle \lim_{a \to \infty} 2\pi \ln a = \infty$, which means that Gabriel's horn has infinite surface area, but a volume of only $\pi$!