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Solid of revolution vs. area below 1/x

sweatingbear

Member
May 3, 2013
91
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
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

sweatingbear

Member
May 3, 2013
91
Thank you both for your replies.