The area of the region bounded by $f(x) = \frac{1}{x}$, $y = 0$, and $x = 1$ is
$$ A = \int_1^{+\infty} f(x) \, \textrm{d}x = \lim_{b \to +\infty} \int_1^b \frac{\textrm{d}x}{x} = \lim_{b \to +\infty} [\ln b - \ln 1] = +\infty $$
The volume of the solid formed by revolving the same region as above is
$$ V = \pi \int_1^{+\infty} [f(x)]^2 \, \textrm{d}x = \lim_{b \to +\infty} \pi \int_1^b \frac{\textrm{d}x}{x^2} = \pi \lim_{b \to +\infty} [-\frac{1}{b} + 1] = \pi $$
I understand that $y = \frac{1}{x}$ and $y = \frac{1}{x^2}$ are different, so it makes perfect algebraic sense that the area under both curves will be different; one being divergent and the other convergent in this question. But how can it make geometric sense that revolving an infinite area results in a finite volume?