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Washer Method

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    y=x^(3) +1, x=1, y=1; rotated about x=-1

    2. Relevant equations

    Washer Method. Pi * Integral from a to b of [Outer radius]^2-[inner radius]^2

    3. The attempt at a solution

    I understand the shell method version but I wanted to learn the washer way for this one.

    Pi* Integral from 1 to 2 of (1+1)^(2)-[1+(y-1)^(1/3)]^(2) dy.
    I am sure this is the wrong way to set it up because I get like pi lol. The answer should be 9pi/10. Like I said I know how to do this through the shells but not the washer. I also do not know how one can see that the shells would be easier to use. Thanks for the help!!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 4, 2012 #2
    You set it up correctly. You must be doing the integration incorrectly.
     
  4. Apr 4, 2012 #3

    SammyS

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    Looks good.

    By the way,
    [itex]\displaystyle \int_1^2 \left(4-(1+\sqrt[3]{y-1}\,)^2 \right) dy=\frac{9}{10}[/itex]​
     
  5. Apr 4, 2012 #4
    Thank you both for helping out :). I got it thanks to you guys. Yay!! I was just integrating wrong. My mistake was that when I squared the 1/3 for some reason I just removed the root instead of making it 2/3 ha. Once again thank you both :)
     
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