Solid of revolution vs. area below 1/x

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Discussion Overview

The discussion revolves around the comparison between the area under the curve of the function $$\frac{1}{x}$$ from $$[1, \infty)$$ and the volume of the solid of revolution generated by rotating this curve about the x-axis. Participants explore the apparent paradox that the area diverges while the volume converges.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions the logic behind the area under the graph of $$\frac{1}{x}$$ being infinite while the volume of the solid of revolution is finite.
  • Another participant reiterates this question and provides calculations showing that the area diverges to infinity, while the volume converges to $$\pi$$.
  • A third participant suggests an external resource, "Gabriel's Horn," which may provide additional context or insight into the topic.
  • One participant expresses gratitude for the responses received, indicating engagement with the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the underlying logic of the situation, and the discussion remains unresolved regarding the paradox presented.

Contextual Notes

The discussion highlights the divergence of the area and convergence of the volume, but does not delve into the underlying assumptions or definitions that may clarify this paradox.

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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