# Solid of revolution vs. area below 1/x

#### sweatingbear

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

#### chisigma

##### Well-known member
How come the area below the graph of $$\displaystyle \frac 1x$$ between $$\displaystyle [1, \infty)$$ 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

$\chi$ $\sigma$