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Simple shell

  1. Feb 21, 2005 #1
    "simple" shell

    I know this is relatively simple, but I'm a little rusty. Could someone help me out? We want to find the volume of the solid obtained by rotating the region bounded by the curves y=x^4 and y=1 about the line y=7 using the cylindrical shell method.

    According to my book the general formula for cylindrical shell method is: V=(circumference)(height)(thickness) or (2pi*r)(r*h)(delta r). So I set up the integral as (2pi) integral [7x -x^5] dx. The boundaries are found by setting x^4=1, which yields -1 and 1. After differentiating we have 2pi[7/2x^2-1/6x^6] from -1 to 1. Because of my boundaries, I initially got the volume=0, but I don't think that's possible. I assumed the minus sign should be a plus, but after adding and multiplying by 2pi, I still got the wrong answer.

    Any clues would be much appreciated! Thanks!
    Last edited: Feb 21, 2005
  2. jcsd
  3. Feb 21, 2005 #2


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    You're rotating the region about the line y=7 right?
    Then if you're using cylindrical shells, you should integrate wtr y (from 0 to 1).
    The radius of the shell is [itex]7-y^{1/4}[/itex].

    Try setting up the integral again.
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