Why Do We Use Frustrums Instead of Disks to Find the Surface of Revolution?

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Discussion Overview

The discussion revolves around the use of frustums versus disks in calculating the surface area of solids of revolution. Participants explore the reasoning behind the preference for frustums in this context and question the applicability of disks for surface area calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why frustums are used for calculating surface areas of revolution instead of disks, which are used for volumes of revolution.
  • Another participant notes that they found references suggesting that disks can be used for surface area calculations, seeking clarification on the differences.
  • A participant mentions that calculations indicate different results when using disks versus frustums due to the differing formulas for surface area.
  • It is suggested that for solids of revolution, the choice between disks and frustums may not significantly impact the results.
  • A later reply introduces an error analysis, indicating that the error in using disks approaches zero as the thickness of the disks decreases, while also noting a specific error related to surface area calculations.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the use of disks for surface area calculations, with some suggesting that frustums are necessary while others propose that disks could suffice. The discussion remains unresolved with competing views on the appropriateness of each method.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations and the specific conditions under which each method may be applicable. The differences in formulas for surface area between disks and frustums are acknowledged but not fully explored.

Bipolarity
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When I learned solids of revolutions, we divided the solid into infinitely thin "disks" and then summed up their individual volumes to get the volume of the surface.

Now in surface of revolutions, we are divided the solid into infinitely thin fustrums and summing up their individual surface area to get the total surface area of the solid.

My question is: Why can't we just use disk to find the surface of revolution by summing up the surface areas of infinitely thin disks? Why use a fustrum for the surface of revolution but not for the solid of revolution?

BiP
 
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Office_Shredder said:
I have never heard the term frustum before so googled it. The notes here
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

seem to suggest that they are doing exactly what you propose. Can you explain the difference?

It seems that surface of revolution only works with fustrum but not with disks.
I have done the calculations, and because the formula for surface of a disk is different from that for a fustrum, the integrals come out different.

Interesting, for solids of revolutions, it doesn't matter whether you use disks or fustrums.

Or maybe I am making a mistake.

BiP
 
Hi Bipolarity! :smile:
Bipolarity said:
Why can't we just use disk to find the surface of revolution by summing up the surface areas of infinitely thin disks? Why use a fustrum for the surface of revolution but not for the solid of revolution?

Because the error in the volume is πx(dx)2tanθ / πx2dx, which -> 0 as dx -> 0, so i's ok to use a disc instead of a frustrum.

(And the error in the area is secθ - 1)
 

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