Shell method versus disk method

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SUMMARY

The discussion centers on the confusion between the shell method and the disk method for calculating volumes of solids of revolution. The specific problem involves revolving the region enclosed by the y-axis, the line y=2, and the curve y=\sqrt{x} around the y-axis. The correct application of the disk method is represented by the integral \(\pi \int y^4 \, dy\) evaluated from 0 to 2, while the shell method should use the integral \(2\pi \int x(2-\sqrt{x}) \, dx\) from 0 to 4. The discrepancy in results arises from the limits of integration used in the shell method.

PREREQUISITES
  • Understanding of calculus concepts, specifically integration
  • Familiarity with the shell method for volume calculation
  • Knowledge of the disk method for volume calculation
  • Ability to interpret and manipulate definite integrals
NEXT STEPS
  • Review the derivation of the shell method for solids of revolution
  • Practice problems involving the disk method with various curves
  • Explore the relationship between different methods of volume calculation
  • Learn about common pitfalls in setting limits of integration
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Students and educators in calculus, mathematicians, and anyone seeking to deepen their understanding of volume calculations using different methods in integral calculus.

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I'm confused when to use the shell method vs. the disk method

Specifically in this problem:

If the region encolsed by the y-axis, the line y=2 and the curve y=[tex]\sqrt{x}[/tex] is revolved around the y-axis, the volume of the solid generated is:

However when I use the shell method and the disk method I get different answers

Can someone tell me what I'm doing wrong?

Disk Method:
[tex]\pi[/tex] [tex]\int[/tex]y^4 dy evaluated from 0 to 2

Shell Method:

2[tex]\pi[/tex][tex]\int[/tex]x(2-[tex]\sqrt{x}[/tex])dx

Help please!
 
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I got the same answer. Remember the second integral is from 0 to 4 not 0 to 2.
 

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