Infinite area under a curve has finite volume of revolution? So I was thinking about the harmonic series, and how it diverges, even though every subsequent term tends toward zero. That meant that its integral from 1 to infinity should also diverge, but would the volume of revolution also diverge (for the function y=1/x)? I quickly realized that its volume is actually finite, because to find the volume of revolution the function being integrated has to be squared, which would give 1/x^2, and, as we all know, that converges. So, my question is, are there other functions that share this property? The only family of functions that I know that satisfy this is 1/x, 2/x, 3/x, etc.

$\frac{1}{x^p}$ with $\frac{1}{2} < p \leq 1$ all satisfy these properties.
Then, by limit comparison test, any positive function $f(x)$ with the propery that there exists a $\frac{1}{2} < p \leq 1$  so that
$$ \lim_{x \to \infty} x^p f(x) = C \in (0, \infty) \,.$$
also has this property... This allows you create lots and lost of example, just add to $\frac{\alpha}{x^p}$ any "smaller" function. (i.e. $o(\frac{1}{x^p} )$)
