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Solids of Revolution - Volume

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the solid obtained by rotating the region bounded by y=2x2-x3 and y=0 about the y-axis


    2. Relevant equations

    There no required method (between Disk, Washer, Shell). In my attempt below, I used the Shell method, I believe. 2π(shell radius)(shell height)dx



    3. The attempt at a solution

    I graphed the equation and ended up with V = ∫20 2π(x)(2x2-x3)dx

    V =2π*limit 0 to 2* (2x3-x4)dx

    V = 2π *limit 0 to 2* x4/2 - x5/5

    V =[tex]2π (1/2 - 1/5)[/tex]
    V= 3π/5

    Am I on the right track or did I mess up horribly in the setup? I feel like the area where I might have made a mistake was the limits I set... and still getting used to the functions on here lol sorry
     
  2. jcsd
  3. Feb 7, 2012 #2

    Dick

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    The setup looks ok to me but it looks like you are putting x=1 for the upper limit of your integral instead of x=2 when you evaluate it. Why would you do that?
     
    Last edited: Feb 7, 2012
  4. Feb 8, 2012 #3
    Ah, just a silly error. So when I plug in 2 for x rather than one. I get 8 - 6.4 = 3.2pi. But other than that it looks alright? I was worried about how I choose to set up my limits since I found the x intercept for the upper limit...but kinda sorta just guessed that the other was 0. There has to be a more technical way to go about that. @_@
     
  5. Feb 8, 2012 #4

    Dick

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    Yes, that's fine. There are two x intercepts. Factor. 2x^2-x^3=x^2*(2-x)=0. So either x=0 or x=2.
     
  6. Feb 8, 2012 #5
    Thanks so much!
     
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