How can Gabriel's Horn have a finite volume?

Question

I’m a highschool student who just finished Calc AB and I’m fascinated by the concept of Gabriel’s Horn but I’m confused by the claim that its volume is finite.

Correct me if I’m wrong but isn’t it somewhat misleading to say that the volume of the horn is finite? Since $\pi$ is irrational and never-ending, the volume wouldn’t be finite, right? Consider this: start with a section of the horn from 0 to some number n in which the volume is exactly 3.14.

$$\int_0^n \pi\left(\frac1x\right)^2 dx = 3.14$$

Then, as you slowly increase the domain of the function to infinity and reveal smaller and smaller cross-sections of the horn, you simply “gather” more and more digits of pi to fill the volume (3.14$\to$3.141$\to$3.1415). So the volume of the horn would always be increasing but at a slower and slower rate. Therefore the limit of the volume would converge to pi as x approaches infinity, but it would not be outright finite as it is always increasing alongside the domain. $$\lim_{x\to\infty} \left[\pi\int_0^x \left(\frac1x\right)^2 dx\right] = \pi$$ Because the limit of the function approaches $\pi$, I think we can just say that the volume is $\pi$ but that doesn't mean that it's finite since $\pi$ by its very nature is an infinite decimal. Is there a mistake in my logic or some part I missed? Any feedback would be helpful.

[Edit: Thank you for all the responses! I think my mistake was that I had the wrong definition of “finite”. I thought that by “finite”, it meant an exact, non-repeating, and rational value (for the volume) as opposed to an infinite value. I failed to realize that although the decimal representation of $\pi$ is infinite, it itself is a clearly defined finite value between 3 and 4. Therefore, the volume of Gabriel’s horn is finite while the surface area isn’t. Math continues to amaze me everyday!]

Answer 1

$\pi$ is a finite number. An infinite number, should it exist, would be a number $y$ for which there is no $x$ such that $y<x$. Clearly $\pi<4$ for example, so $\pi$ is not infinite.

$\pi$ has an infinite number of digits, but so does $1/9=0.11111....$, and it's not rational, but neither is $\sqrt2$, and that also is clearly not infinite.

So, no, the volume is not infinite. You might want to look up "Zeno's paradox".

Answer 2

So the notion of Gabriel's Horn is that the horn itself is a solid of revolution about the x-axis of the function $f(x)=1/x$ over the domain $x\in[0,\infty)$.

The general integral for this volume is:

$$ V =\iiint_\text{Gab's Horn}dxdydz =\int_{x=1}^{\infty}\pi R^2(x)dx =\pi\int_1^\infty x^{-2}dx =\pi[x^{-1}]_\infty^1 =\pi\left[\frac{1}{1}-\frac{1}{\infty}\right] =\pi $$

Which is correct as you claim. This has nothing to do with $\pi$ in and of itself and the fact that $\pi$ appears in this equation is an artifact of the fact that the cross-section being integrated over is circular in nature. Had it been a different solid formed by a different function then we would get a different answer.

Consider the portion of Gabriel's Horn that corresponds to the domain $x\in[\pi,\infty)$. The volume of this portion is then

$$ V=\int_{x=\pi}^{\infty}\pi R^2(x)dx =\pi\left[\frac{1}{\pi}-\frac{1}{\infty}\right] =\frac{\pi}{\pi}=1 $$

The above result is equivalent to cutting off the lower portion of Gabriel's Horn. Clearly $\pi$ has nothing to do with the volume itself from any other perspective than 'just geometry'. The fact that $\pi$ is irrational is irrelevant, and the need to express $\pi$ with infinite digits is true in base-10 arithmetic, $\pi=3.1415926\cdots_{10}$, but in base-$\pi$, we have $\pi=1_\pi$. So the infinite-digit-representation of $\pi$ isn't even generally true in general.