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Disk/Shell Method and Volume

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    Set up the integral (do not evaluate) to find the volume by revolving the region enclosed by y=x^2-2x+2 and y=-x^2+6 about a) x = 3 and b) y = -5.

    2. Relevant equations

    Shell Method: V = 2∏ ∫ (radius)*(height)dx
    Washer Method: V = ∏ ∫ (R^2 - r^2)dx


    3. The attempt at a solution

    I believe I need to use the Shell Method for part A.
    V = 2∏ ∫ (-1→2) (3-x)*(-x^2+6-x^2+2x-2)dx
    V = 2∏ ∫ (-1→2) (3-x)*(-2x^2+2x+4)dx

    I believe I need to use the Disk/Washer Method for part B.
    V = ∏ ∫ (-1→2) [(-x^2 +6+5)^2 - (x^2-2x+2+5)^2]dx
    V = ∏ ∫ (-1→2) [(-x^2 +11)^2 - (x^2-2x+7)^2]dx

    But I'm not sure. Rotating about the axes are a lot simpler and I'm not sure about how to handle the values for x < 0. In my head, I think the equations will handle that for me, but I'm concerned.

    Thanks for the help.
     
  2. jcsd
  3. Oct 12, 2011 #2
    I don't see anything wrong with the set up, assuming I didn't miss anything
     
  4. Oct 12, 2011 #3

    SammyS

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    Looks good to me too.
     
  5. Oct 13, 2011 #4
    Well, look at me go!

    (woot woot)

    Thanks for putting some eyeballs on this me.
     
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