Surface area of a solid of revolution

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SUMMARY

The discussion focuses on calculating the surface area and volume of a solid of revolution formed by revolving the function f(x) = 1/x around the x-axis, specifically for the interval 1 ≤ x < ∞. Participants confirm that the volume of the solid is finite, calculated as π, while the surface area is infinite. This paradox raises questions about the validity of the approach used, prompting critical analysis of the implications of having a finite volume with an infinite surface area.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of solids of revolution.
  • Knowledge of improper integrals and their applications.
  • Basic grasp of surface area and volume calculations.
NEXT STEPS
  • Study the method for calculating surface area and volume of solids of revolution using the disk and washer methods.
  • Learn about improper integrals and their convergence properties.
  • Explore the implications of infinite surface areas in mathematical modeling.
  • Investigate the relationship between geometric properties and physical interpretations in calculus.
USEFUL FOR

Students and educators in calculus, mathematicians exploring geometric properties, and anyone interested in the paradoxes of mathematical analysis related to solids of revolution.

madgab89
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Homework Statement



Consider now a shape that is obtained by revolving
the “infinitely long” function
f (x) = 1/x , 1 ≤ x < ∞
around the x-axis. Find both the surface area of the resulting object, and the
enclosed volume of it, i.e. the volume of the solid obtained from revolving the
area under the graph of f (x) around the x-axis.

If you do this correctly, you will find an infinite surface area but a finite volume. This would mean that a certain (finite) amount of paint is sufficient to fill up our object, but that this
amount would not be enough to actually paint it. Does this surprising result
mean that there is something fundamentally wrong with our approach? Why
or why not? Comment critically!

Homework Equations


I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.
 
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madgab89 said:

Homework Statement



Consider now a shape that is obtained by revolving
the “infinitely long” function
f (x) = 1/x , 1 ≤ x < ∞
around the x-axis. Find both the surface area of the resulting object, and the
enclosed volume of it, i.e. the volume of the solid obtained from revolving the
area under the graph of f (x) around the x-axis.

If you do this correctly, you will find an infinite surface area but a finite volume. This would mean that a certain (finite) amount of paint is sufficient to fill up our object, but that this
amount would not be enough to actually paint it. Does this surprising result
mean that there is something fundamentally wrong with our approach? Why
or why not? Comment critically!

Homework Equations


I understand how to calculate the surface area and the volume of a solid of revolution, however it's the second part of the question I don't really undertand, the part about there being something fundamentally wrong with our approach. I was wondering if anyone could give me some hints to point me in the right direction.

A good start would be to actually calculate the surface area and volume for this solid. Keep in mind that you are going to have improper integrals for both, since one of the limits of integration is \infty.
 
Yeah, I did that:

I got pi as the volume of the solid and undefined (infinity) for the surface area. However its the theoretical part that I'm unsure about, where its asking if there's something wrong with the method...
 

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