Subject: Calculus

# Gabriels Horn

Gabriel’s Horn in calculus describes the graph of a function f(x) = 1/x rotated about the x-axis. It has a finite volume but has an infinite surface or in analogy, one can fill a Gabriel’s horn with paint but one cannot paint its surface itself. And how is that possible?

## Formation

Take into consideration the graph of f(x) = 1/x where x > 0 (See figure below). We can see that this function is asymptotic when x = 0.

For simplicity, let us consider values of x from 1 to a, where a > 1. Then we rotate the curve about the x-axis. This would result to solid that looks like a horn as shown in figure below. This is what we now call as the Gabriel’s horn which was named after the Archangel Gabriel who was said to blow the horn to mark the beginning of the Judgment Day. Gabriel’s Horn is also known as Torricelli’s Trumpet which was named after an Italian physicist and mathematician who first studied the properties of this function.

Going back to the question how is it possible that it encloses a finite volume yet at the same time having an infinite surface, we can answer this mathematically. Let us first calculate the volume of the figure using Disk method since we are to integrate along the axis of revolution (x-axis). Our volume V can then be expressed as,

V=\pi\int_{1}^{u}\frac{1}{x^2}dx=\pi\left (1-\frac{1}{a}\right )

As can be inferred from the equation, if as a approaches infinity, 1/a approaches zero. Simply put, The volume of the figure would never exceed π as a approaches infinity. We can further illustrate this one by denoting it with limits,

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

This clearly indicates that the volume, having a finite limit, is also finite.

The area is given by the integral

A=2\pi\int_{1}^{a}\frac{\sqrt{1+\frac{1}{x^4}}}{x}dx=2\pi\int_{1}^{a}\frac{\sqrt{1}}{x}dx=2\pi \ln{a}

Taking the limit of A, we will have

\lim_{a \to \infty}(2 \pi \ln{a})=\infty

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