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Need Help With This Integration

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Hey guys, I am working an integration problem and am having a hard time.
    This is the problem : Let R be the region bounded by Y = 2x^(-3/2), Y = 2, Y = 16, and X = 0. Use the shell method to find the volume of the solid generated when R is revolved about the X-Axis.

    The radius is just Y since it's about an axis.
    The bounds are from 2 to 16.
    I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
    So wouldn't my integral be 2∏∫(Y)((1/2)Y^(-2/3)dy from 2 to 16?
    the anser is 90∏, and that is not what I am getting :/



    2. Relevant equations

    So I know the shell method is 2∏∫(radius)(height).
    Since its about the X-Axis, we are integrating with respect to Y, which means we have to solve for X.





    3. The attempt at a solution

    The radius is just Y since it's about an axis.
    The bounds are from 2 to 16.
    I don't know if I am solving for X wrong, when I solve for X, I get X = (1/2)Y^(-2/3)
    So wouldn't my integral be 2∏∫(Y)((1/2)Y^(-2/3) from 2 to 16?
    the anser is 90pi, and that is not what I am getting :/
     
    Last edited: Feb 16, 2014
  2. jcsd
  3. Feb 16, 2014 #2

    SammyS

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    Hello Shakenbake158. Welcome to PF.

    You're either solving for x incorrectly, or you're incorrectly expressing the result.

    (y/2) should be raised to the -2/3 power, not y alone.
     
  4. Feb 16, 2014 #3

    Thank you for the greeting!

    I thought that I might be solving for X wrong.

    So now I have 2∏∫(y)((y/2)^(-2/3))dy

    My algerbra is failing me, how do I multiple these two terms?
     
  5. Feb 16, 2014 #4

    SammyS

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    [itex]\displaystyle \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}[/itex]


    [itex]\displaystyle a^m a^n=\left(a\right)^{(m+n)}[/itex]
     
  6. Feb 16, 2014 #5

    Thank you so much for the help. I was finally able to get 90∏.

    I was just making algebraic mistakes.

    I appreciate the help.
     
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