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Volumes of revolution not around the axis

  1. May 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the volume of [itex] y = 2x^2 [/itex] y = 0, x = 2 when it is revolved around the line y = 8.


    2. Relevant equations
    Integral formulas for volumes by discs, washers and cylinders.


    3. The attempt at a solution
    Translate the curve so that axis of revolution is along the X axis. Is this the right idea? This gives [itex] y = 2x^2 - 8 [/itex] . I would integrate this and subtract from the volume of the cylinder with radius 8 and height 2:

    [tex] \pi(8^2)(2) - \int_0^2 \pi(2x^2 - 8)^2\,dx [/tex]

    Is this the right approach?

    Thanks,
    Sheldon
     
    Last edited: May 21, 2007
  2. jcsd
  3. May 21, 2007 #2

    Dick

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    This looks correct to me.
     
  4. May 21, 2007 #3
    Thanks Dick, I really appreciate it.
     
  5. May 21, 2007 #4

    orb

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    [tex]\pi\int_0^2 [ (8)^2 - (8 - 2x^2)^2 ] \,dx[/tex]

    I think that works, because if you use the washer method, the outer radius is just the part that has a y-length of 8, and the inner radius is the part above the function and under y=8, so using pi (R^2 - r^2) integrated, that's what I get. Hope that helps :)
     
  6. May 21, 2007 #5

    Dick

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    It's the same thing he already wrote.
     
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