MHB Solid of revolution vs. area below 1/x

SweatingBear
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How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.
 
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sweatingbear said:
How come the area below the graph of $$\frac 1x$$ between $$[1, \infty)$$ does not exist, but the solid of revolution below the same graph in that same interval does exist? I do not see the logic.

The calculus demonstrates that the area... $\displaystyle A = \int_{1}^{\infty} \frac{dx}{x} = \infty$ (1)

...and the volume of its the solid of revolution is finite and is... $\displaystyle V= \pi\ \int_{1}^{\infty} \frac{dx}{x^{2}} = \pi$ (2)

It may seem incredible but this 'paradox' was discovered by Pappo di Alessandria in the fourth century after Christ [!]...Kind regards $\chi$ $\sigma$
 
Thank you both for your replies.
 
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