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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?

Chewers Jingoist
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    Why are you not surprised that the area under the line of infinite length can be finite? This is the same thing – Yuriy S Feb 02 '16 at 19:03
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    @Yuriy S: I've also wondered about this for many years (decades, in fact). For some reason the one-and-two dimensional version seems to cause MUCH less concern than the two-and-three dimensional version. – Dave L. Renfro Feb 02 '16 at 19:12

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