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Volume by Shells

  1. Dec 5, 2006 #1
    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:
     
  2. jcsd
  3. Dec 5, 2006 #2

    quasar987

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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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