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Find the volume of the solid obtained by revolving the region

  1. Apr 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Alright well I think I have this right, but my teacher has it done a different way..

    Find the volume of the solid obtained by revolving the region bounded by y=x2+1; y=9-x2 about y=-1.

    2. Relevant equations



    3. The attempt at a solution
    So I thought the integral would be:
    [tex]2\pi\int_{0}^{2}(9-x^{2})^{2}-(x^{2}+3)^{2}dx[/tex]
    I thought it was x2+3 because you have to rotate it around y=-1 so the inner radius would be 2 added to the 2nd equation.
    Oh yeah and I just did the volume from 0 to 2 and multiplied by 2..

    Thanks for any help.
     
  2. jcsd
  3. Apr 22, 2013 #2

    Dick

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    That's totally misguided. The inner radius is only 2 at x=0, isn't it? Did you draw a sketch? At an arbitrary value of x the inner radius is (x^2+1)-(-1), isn't it? What's the outer radius?
     
    Last edited: Apr 22, 2013
  4. Apr 22, 2013 #3
    y=9-x2-(-1) so 10-x2 for the outer radius?

    But either way his answer is wrong he didn't even take that into account.
     
  5. Apr 22, 2013 #4

    Dick

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    If that's the way your teacher did it, then it looks wrong to me. I thought you were showing the way you did it?
     
  6. Apr 22, 2013 #5

    Curious3141

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    The steps in solving these problems are very standard:

    1) Sketch the curves and highlight the region of interest. Draw in the axis of rotation.

    2) Write an expression for the element of volume.

    3) Integrate and evaluate the definite integral.

    Try going through the steps systematically.
     
  7. Apr 22, 2013 #6
    He just subtracted the first equation (squared) from the second equation (squared).
    I thought that he did something wrong with it, I don't think he took into account the y=-1 part.
     
  8. Apr 22, 2013 #7

    Curious3141

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    If he'd added one to each of the y-expressions before squaring then subtracting, then he's taken the axis into account.
     
  9. Apr 22, 2013 #8

    Dick

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    So what did he show? What you showed is not just subtracting the first equation squared from the second squared. You magically added two to just one of them. Why?
     
    Last edited: Apr 22, 2013
  10. Apr 22, 2013 #9
    No he just left them as they are.
     
  11. Apr 22, 2013 #10

    Dick

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    Then that would correspond to a rotation around y=0. I don't think that's what you want.
     
  12. Apr 22, 2013 #11
    Ok that's what I thought, thanks lol.
     
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