Intuition behind this interesting calculus result?

Yes, this was considered a pretty strange phenomenon when Torricelli first constructed such an example in 1643. (Torricelli himself found it so incredible that he offered two different proofs that the volume of his shape was finite, the better to convince himself and the reader).

I think there are two main mental hurdles that one must conquer in order to shake the impression that this is paradoxical:

• An infinite sequence of positive numbers can have finite sum.

• Volumes of revolution with the same finite cross-sectional area can have as large or small volume as you want.

The first of these has occupied minds since Zeno's paradoxes. Most of us are eventually able to come to sort of peace with it, if only by bludgeoning ourselves with the many examples of it.

For the second, just compare the volumes of cylinders you get by rotating a rectangle of unit area around one of its sides. The sides of the rectangle is $a$ by $\frac1a$ for some $a$, and calculating its volume we get $\pi a^2\cdot\frac1a = \pi a$ which we can make anything we want just by choosing $a$ appropriately. What's going on is that a small piece of area close to the rotation axis contributes less volume than the same area longer away, so by choosing a long thin rectangle that hugs the axis tightly, we can make the volume as small as we want to.

Putting these two together: Choose a convergent series of positive numbers $$ b_1 + b_2 + b_3 + \cdots = c < \infty$$ and for each $b_i$ pick the cylinder with volume $b_i$ and cross-sectional area $1$. Stack all of these cylinders after one another, and you get an infinitely long volume of revolution with infinite cross-section but finite volume, namely $c$.

This is perhaps not as beautiful as your nice smooth $\frac{\log x}{x}$, but is arguably easier to wrap one's head around because one needs only deal with discrete series rather than continuous integrals.

Yes, it is an example that shows that we cannot assume that ''an infinite area rotated around the x axis would produce an infinite volume''.

Intuitively this is because the area goes to infinity because $f(x)=\frac{\ln x}{x}$ goes to $0$ too slowly, but the volume $f^2(x)=\left(\frac{\ln x}{x}\right)^2$ goes to $0$ more quickly.

I like to think of it by comparing it to what you do to convergent improper integrals. The length of the curve is infinite but the area under the curve is convergent and infinite. You can apply the same intuition in three dimensions. I hope this helps.

You'd assume an infinite area rotated around the x axis would produce an infinite volume... right?

Since you asked for intuition, here's a non-technical intuition for Gabriel's Horn (a simpler, but essentially the same, case):

When you integrate to get the area, you're summing up a bunch of vertical lines, more or less. Call the height at $x$ the radius $r(x)$. When you integrate to get the volume, you're summing up a bunch of circles, each of area proportional to $r(x)^2$. Then recall that, while $1/x$ is a divergent sequence, $1/x^2$ is convergent.

In short, the area of the circles you're summing for volume shrinks faster than their radii you're summing for area, because for large $r$, $1/r~~>~~1/r^2$.

Suppose a solid of revolution has infinite surface area but finite volume. Imagine filling it with a hypothetical paint that, unlike real paint, isn't made of particles. Although the shape becomes arbitrarily thin as paint flows through it, the special paint we're using never gets stuck in a narrow path. Clearly, only a finite volume is needed to fill the solid, thereby painting its inner surface. However, the fact that the surface area is infinite implies that, if you instead paint the outside, a finite volume of paint has to be infinitely thin to suffice. In other words, neither the inner nor outer surface can be painted with realistic paint, but both can be with the hypothetical paint. The interpretation is that $0\times\infty$ is an indefinite form (rather than being infinite).

Let's consider a simpler geometry. The cross-sectional area and volume of a cylinder.

The volume of a cylinder is $\pi r^2 h$, where $r$ is the radius and $h$ is the length or height of the cylinder. Let's fix the volume at $\pi r^2h=1$, or $h=\frac1{\pi r^2}$.

Now, the cross-sectional area is given by $2rh=\frac2{\pi r}$. As a result, as we can choose $r$ as small as we want, we can make the cross-sectional area as large as we want. This does not change the volume.

The result in the question works similarly, it simply avoids the problem of the limit to infinity (if you take the limit as $r\to\infty$, you end up with a line rather than a cylinder) by having the radius vary along it.

For a very simple intuition, consider the square-cube law. In general, rescaling an object by a factor of $\ell$ changes the volume by a factor of $\ell^3$, while the surface area and cross sectional area scales by a factor of $\ell^2$.

Square-Cube for a Ball

For example, consider a three-dimensional ball (a solid sphere). For a fixed radius $r$, it has ...

• A volume of $V(r) = \frac{4}{3} \pi r^3$.
• A surface area of $S(r) = 4 \pi r^2$
• A cross sectional area (area of slice through the middle) of $A(r) = \pi r^2$

Consider the following limit: $$\lim_{r \to \infty} \frac{V(r)}{S(r)} = \lim_{r \to \infty} \frac{1}{3} r = \infty. $$ So as $r \to \infty$ we gain volume much faster than we gain surface area.

However, consider what happens when $r$ gets really small: $$\lim_{r \to 0^+} \frac{V(r)}{S(r)} = \lim_{r \to 0^+} \frac{1}{3} r = 0. $$ This tells us that we have a miniscule amount of volume for a small surface area---in the limit sense, we lose volume much faster than we lose surface area as the radius decreases.

"Square-Cube" for Riemann Sum

This same sort of reasoning can be extended to looking at volumes of revolution, approximated as a Riemann sum. Breaking the interval of integration into boxes of width one, we can approximate the volume of revolution as cylinders along the $x$-axis of length $1$. Each of these cylinders has a...

• Volume of $V(r) = \pi r^2 \cdot 1$
• Surface area of $S(r) = 2 \pi r \cdot 1$
• Area under the curve of $A(r) = r \cdot 1$

Since $$\lim_{r \to 0^+} \frac{V(r)}{S(r)} = \lim_{r \to 0^+} \frac{1}{2} r =0,$$ we have that we lose volume faster than we lose surface area. Similarly, $$\lim_{r \to 0^+} \frac{V(r)}{A(r)} = \lim_{r \to 0^+} \pi r =0.$$ Thus, as the height of our function ($r = r(x)$) goes to zero, the volume of revolution on each interval goes to zero even faster.

This all really comes down to the fact that if $0 < r < 1$, then $0< r^2 < r < 1$. In fact, $0< \dots <r^3 < r^2 < r < 1$.

Note that, for instance, $$\int_1^\infty\frac1 {x^2}\,dx <\infty$$ so the area below the curve is finite, while the length of the curve is infinite.

What is happening is simply that the length does not increase as fast as the height shrinks.

Similarly, in your example, the area does not increase as fast as the diameter decreases.