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Simple volume help please

  1. Mar 8, 2005 #1
    Ok,so I have a simple volume problem that i'm having a little trouble with.

    I want to rotate the region in the first quadrant bounded by the following formulas around the Y axis.
    [tex]
    y=x^3[/tex]
    [tex]y=2x-x^2[/tex]

    I solved for x and got
    [tex]
    x=\sqrt[3]{y}[/tex]
    [tex]x=1- \sqrt{1-y}[/tex] I know I get two roots but I am pretty sure this is the one I want.

    I do have one question here, when given such equations as [tex]y=2x-x^2[/tex] how do you go about solving for x? I used Mathematica because I couldnt get it, but anyways back to the main problem.

    so I then did [tex]\pi \int^1 _0 \sqrt[3]{y}^2 - (1- \sqrt{1-y})^2 dy[/tex]

    I expanded [tex](1- \sqrt{1-y})^2[/tex] and then integrated the expresion and got the following:

    [tex]\pi [\frac{3y^\frac{5}{3}}{5} - 2y - \frac{4}{3} (1-y)^{\frac{3}{2}}+\frac{y^2}{2}]^1_0[/tex]

    This isn't right however and I assume I messed up somewhere in my final integration, any help? Thanks a lot. I am also going to assume it has to do with [tex](1- \sqrt{1-y})^2[/tex], I did U substitution on the term [tex]2 \sqrt{1-y}[/tex] and am thinking I messed that up, but once again, am not sure. The only other thing I can think about is maybe I missed up my functions somehow. Thanks for any help.
     
    Last edited: Mar 8, 2005
  2. jcsd
  3. Mar 8, 2005 #2

    Galileo

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    It's a quadratic in x:

    [tex]x^2-2x+y=0 \Rightarrow x=1\pm \sqrt{1-y}[/tex]
     
  4. Mar 8, 2005 #3

    dextercioby

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    At a first glance,your integration seems okay.

    Daniel.
     
  5. Mar 8, 2005 #4
    Oh, sorry, I think I just messed up evaluating it, I forgot that all the terms aren't 0 when it is evaluated at 0...I make the dumbest mistakes! :blushing:
     
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