Using the Shell method to find the volume of a solid

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SUMMARY

The discussion focuses on using the shell method to calculate the volume of a solid generated by revolving the region bounded by the curves x=2√y, x=-2y, and y=1 around the x-axis. The initial volume equation presented is v=∫(from 0 to 2) 2π(y-1)((2√y)-(-2y)), which is identified as incorrect. Participants question the choice of (y-1) for the radius and the limits of integration, prompting a reevaluation of the setup for accurate volume calculation.

PREREQUISITES
  • Understanding of the shell method in calculus
  • Familiarity with integral calculus and volume calculations
  • Knowledge of the curves x=2√y and x=-2y
  • Ability to graph functions and identify bounded regions
NEXT STEPS
  • Review the shell method for volume calculations in calculus
  • Learn about setting up integrals for revolving solids
  • Explore the concept of limits of integration in volume problems
  • Practice graphing functions to identify bounded regions for solids of revolution
USEFUL FOR

Students and educators in calculus, particularly those focusing on volume calculations using the shell method, as well as anyone seeking to improve their understanding of integral calculus applications.

Zuni Tiberius
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Use the shell method to find the volume of a solid generated by revolving the region bounded by the given curves and lines about the x-axis.

x=2√y
x=-2y
y=1

So I drew a graph and then using the equation v=∫2πrh

and I got the following

v=∫(from 0 to 2) 2π(y-1)((2√y)-(-2y))

but this is wrong. any suggestions?
 
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Zuni Tiberius said:
I got the following

v=∫(from 0 to 2) 2π(y-1)((2√y)-(-2y))

but this is wrong. any suggestions?
Why do you have (y-1) for the radius? And why are your limits 0 to 2?
 

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