Opposite of Gabriel's Horn Paradox

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The discussion centers on the mathematical exploration of the integral ∫ from 0 to 1 1/x²/³ dx, which evaluates to 3, and its implications when rotated about the x-axis, resulting in a divergent volume of π * ∫ from 0 to 1 1/x⁴/³ dx. Participants clarify that while Gabriel's Horn illustrates an infinite area yielding a finite volume, the converse—an infinite volume with a finite area—is not valid. The conversation highlights the distinction between "area under the curve" and "surface area of the solid of revolution," emphasizing the need for precise calculations in these contexts.

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  • Familiarity with the concept of solids of revolution in calculus.
  • Knowledge of the Painter's Paradox and its implications in geometry.
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  • Study the implications of the Painter's Paradox in various mathematical contexts.
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fakecop
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We are all familiar with Gabriel's Horn, where the function f(x) = 1/x generates an infinite area but a finite volume when revolved around the x-axis.

So the other day I stumbled upon a particular interesting integral: ∫ from 0 to 1 1/x2/3 dx

Instead of infinite limits of integration, we have an infinite integrand. evaluating this integral, we have 3.

When we rotate the integral about the x-axis, however, we have pi* ∫ from 0 to 1 1/x4/3 dx, which diverges to infinity.

Is this possible? I know that an infinite area can produce a finite volume of revolution, but can the converse of the statement be true? or have I done something wrong?
 
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SteamKing said:
http://en.wikipedia.org/wiki/Gabriel's_Horn

Gabriel's horn is an example of the Painter's Paradox. According to the article above, the converse (infinite volume, finite area) is not true.

I don't think you are properly calculating the surface area or the volume of the your example function.

See: http://en.wikipedia.org/wiki/Pappus's_centroid_theorem

I think the OP is using "area" in the sense of "area under the curve", not "surface area of the solid of revolution", which is what the Painter's Paradox is concerned with.
 
Yes, sorry for not clarifying but by area I meant "area under the curve", which is a different calculation than "surface area of the solid of revolution". With solids of revolution that are infinitely long, I cannot find a connection between the surface area and the area under the curve.
 
Very interesting observation. I believe you are correct in your finding. Thank you for sharing this insight.

Junaid
 

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