Reverse of Gabriel's Horn/Toricelli's Trumpet - need explanation of the proof I am writing an exploration on the Gabriel's Horn (Toricelli's Trumpet) and as one of my chapters I wanted to take a look at the existence of a solid of revolution with infinite volume and finite surface area. The exclusive proof I found is 



My problem is that I don't understand what actually happens there, as I am not familiar with the sup notation. Moreover, I don't know where the $f(x)^2 -f(1)^2$ comes from, nor why it can be transformed to an integral and later to an integral of a derivative. Why are the limits of the integral modified and where the right side of the inequality comes from? Eventually, I don't understand what $M$ means in the volume formula. It is surely beyond my current level of maths (final high school grade), and I couldn't find any proper source to understand the application of those lim sups in there. I would be grateful for a step by step  explanation, assume that I will understand basic calculus and standard limits, but nothing beyond.
 A: Sorry to take so long to get back, but while I get the idea the author was going for in that proof, it has so many problems. That first inequality chain - whose purpose is just to show that $f(x)$ is bounded - is broken, and I haven't figured out how to fix it. Perhaps that ridiculous statement is actually an artifact of sloppy mathematics, not sloppy translation, after all.
Besides which, this theorem is only about solids of revolution, which is just one small class of the solids you need to be concerned with. Consider a similarly shaped solid, with with square cross-sectional areas instead of circular. A minor change, but the theorem doesn't apply any longer, even if the proof were fixed.
So let me provide the outline of a different, more general, proof. Using cylindrical coordinates, consider a solid whose surface is given by a continuously differentialble function $r = R(\theta, z)$ with $R(0, z) = R(2\pi, z)$. And for convenience, I will also assume that there is a value $M$ such that if $z \ge M$, then $|R(\theta,z)| < 1$ for all $\theta$. Effectively, this means that if you go out far enough on the $z$-axis, the solid will be contained within a cylinder of radius $1$ from then on. (This assumption is not actually needed, but the proof becomes much more complicated without it, and I think you'll understand it better without the added complication.)
The differential of volume for cylindrical coordinates is $dV = r\,dr\,d\theta\,dz$. The differential of surface area is $dA = ds\,dz$, where $ds$ is arclength along the curve $\theta \mapsto R(\theta, z)$ for fixed $z$. So $$ds = \sqrt {dR^2 + R^2d\theta^2} = \left(\sqrt{\frac{\partial R}{\partial \theta}^2 + R^2}\right)d\theta$$
So, surface area is given by $$A = \int dA = \int_1^\infty \int_0^{2\pi} \sqrt{R_\theta^2 + R^2}\,d\theta\,dz$$
And volume is given by $$V = \int dV = \int_1^\infty \int_0^{2\pi}\int_0^{R(\theta,z)} r\,dr\,d\theta\,dz = \frac12\int_1^\infty \int_0^{2\pi} R^2\,d\theta\,dz $$
Now we can break the area and volume up by the plane $z = M$: $A = A_1 + A_2, V = V_1 + V_2$, where
$$A_1 = \int_1^M \int_0^{2\pi} \sqrt{R_\theta^2 + R^2}\,d\theta\,dz,\qquad A_2 = \int_M^\infty \int_0^{2\pi} \sqrt{R_\theta^2 + R^2}\,d\theta\,dz\\V_1 =\frac12\int_1^M \int_0^{2\pi} R^2\,d\theta\,dz, \qquad V_2 = \frac12\int_M^\infty \int_0^{2\pi} R^2\,d\theta\,dz$$
$A_1$ and $V_1$ are the integrals of bounded functions over bounded regions, and therefore must be finite. For $A_2$ and $V_2$, we know that $R < 1$, and therefore $R^2 < |R|$. But then
$$\sqrt{R_\theta^2 + R^2} \ge \sqrt{R^2} = |R| > R^2 \ge \frac{R^2}2$$
From which it follows that $A_2 \ge V_2$.
If $V$ is infinite, then since $V_1$ is finite, it must be that $V_2$ is infinite. But that means $A_2$ is also infinite, and so is the full surface area $A$.
By the contrapositive, if $A$ is finite, the $V$ must be finite as well.
