Volume by Shells: Find Length of Radius?

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SUMMARY

The discussion focuses on calculating the volume of a solid generated by revolving the region bounded by the equations y=1/(x^2), y=0, x=1, and x=4 around the y-axis using the shell method. The radius of the shell is confirmed to be x when integrating over x, leading to the volume formula V=∫14 (2πx)(height of tube) dx. The main challenge lies in determining the height function h(x) for this integration method. Alternatively, integrating over y requires finding a disk area function A(y) for the volume calculation.

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  • Understanding of integral calculus and volume calculations
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  • Knowledge of functions and their graphs, specifically y=1/(x^2)
  • Ability to manipulate and evaluate definite integrals
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  • Study the shell method for volume integration in detail
  • Learn how to derive height functions for various solid shapes
  • Explore disk and washer methods for volume calculations
  • Practice problems involving volume of revolution around different axes
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Students and educators in calculus, particularly those focusing on volume calculations using the shell method, as well as anyone seeking to deepen their understanding of integration techniques in solid geometry.

doppelganger007
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Hi,

I'm doing a problem about finding volume using shells, and I'm really confused about what the the width function should be. The question asks to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated line:

y=1/(x^2), y=0, x=1, x=4, and revolve that region around the y-axis.

Wouldn't the length of the radius just be x? That's what I think it is but I'm not sure. Help would be appreciated, thanks :smile:
 
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There are two ways of integrating using shells: integrating over x or over y (adding up tubes of adding up disks).

If you want to integrate over x, then yea, the radius is just x and the volume is

[tex]V=\int_1^4 (2\pi x)(\mbox{height of tube})dx[/tex]

So the "challenge" with the tube method is really more of finding the height function h(x).

If you want to integrate over y then it's about finding a "disk area function" A(y) such that

[tex]V=\int_{1/4^2}^{1/1^2}A(y)dy[/tex]
 

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