$\frac1x$ rotational volume vs area

Question

The area under the function $f(x) = \frac1x$ in the interval $(1, \infty)$ diverges while the rotational volume converge to $\pi$.

Question: How can this be explained without the use of integration to develope a deeper and better understanding?

Answer 1

We often think of squaring a number as making it bigger. $2^2 = 4 > 2$, $3^2 = 9 > 3$, etc. However, for numbers smaller than $1$, squaring them makes them smaller. $(1/2)^2 = 1/4 < 1/2$, $(1/3)^2 = 1/9 < 1/3$, etc.

For $x \in (1,\infty)$, $1/x < 1$ so $(1/x)^2 < 1/x$. So the cross-sectional area $\pi/x^2$ of the volume of rotation of $1/x$ becomes much much smaller than $1/x$ as $x \to\infty$.

The divergence of the integral

$$ \int_1^\infty \frac{1}{x} \, dx $$

is actually quite special. It is a very slowly diverging integral, with $\int_1^a 1/x \, dx = \ln(a)$ which grows to infinity quite slowly. So by squaring $1/x$, we make it go to zero much faster than $1/x$ which gives us convergence.

This gives you an intuitive justification for why you shouldn't be surprised that volume integral converges. I am not aware of a formal proof that is not harder then actually evaluating the integral.