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Solid Of Revolution Problem - Washer

  1. Mar 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the volume of a solid created when the area between the function y=x^2+1 and the x-axis (for 0<x<2) is rotated about the line y=-2.

    2. Relevant equations

    Vs = ∏*r^2*h

    3. The attempt at a solution

    I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

    I think this creates a full solid around y=-2 -- would I just want to add 1 to the radius instead of 3?

    Many thanks!
     
  2. jcsd
  3. Mar 22, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Bellwether! Welcome to PF! :smile:

    (try using the X2 button just above the Reply box :wink:)
    That's for the outer radius …

    you now need to subtract the volume for the inner radius. :wink:
     
  4. Mar 22, 2012 #3
    For problems like this, I used to prefer adding 2 to the function to shift the whole thing up 2 units of y. Then you can simply revolve the function around the line y = 0, and the volume result will be the same.
     
  5. Mar 22, 2012 #4
    Thanks for the welcome!

    So it's, ∫∏((x2+1)+3)2 - 22) ?
     
  6. Mar 23, 2012 #5

    tiny-tim

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    Hi Bellwether! :smile:

    (just got up :zzz:)
    ∫ π((x2+1) + 2)2 - 22) dx :wink:
     
  7. Mar 23, 2012 #6
    I'm confused -- why is the outer radius just +2 , instead of 3? Isn't (x2 + 1) three units away from the axis of revolution?
     
  8. Mar 23, 2012 #7

    tiny-tim

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    no, you've counted the "1" twice …

    (x2 + 1) - (-2 ) = x2 + 3 :wink:
     
  9. Mar 23, 2012 #8
    Of course, thank you.
     
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