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Volume of Revolution question

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A solid is generated by revolving the region bounded by y = x2/2 and y = 2 around the y-axis. A hole, centered along the axis of revolution, is drilled through this solid so that one-fourth of the volume is removed. Find the diameter of the hole.


    3. The attempt at a solution
    I'm going with cylindrical shells this time around. I'm integrating from x = 0 to x = 2. I think my height is 2-x and my radii are all going to be generated by the function x2/2. After integration, I get the overall volume is to be 4pi/3. If one quarter of that is taken out after drilling this hole, I have the volume of this figure to be pi. I believe it's a right cylindrical shaped whole, so the volume of a cylinder is pir2h. I need the radius of one of these cross sections so I need some way to relate the volume. Or maybe I don't, this is where I need assistance. Hopefully i'm not over-thinking this. Thanks.
     
  2. jcsd
  3. Feb 26, 2013 #2
    I believe i've made a mistake on the radius / height. Radius = x and Height = x2/2, after integration I get vt = 4pi so the volume of the cylinder is 3pi.
     
  4. Feb 26, 2013 #3

    haruspex

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    Yes, except that the removed core is not exactly a cylinder. What do you get for the diameter of the hole?
     
  5. Feb 26, 2013 #4
    I don't know how to find it, especially since I don't know what the figure of the hole is.

    Edit: the diameter seems like it would be 2(y/2)1/2 but that's sort of a shot in the dark.
    Edit: that's wrong, nvm.
     
    Last edited: Feb 26, 2013
  6. Feb 26, 2013 #5

    SammyS

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    The height of each shell should be 2 - x2/2 .

    Do the same integral but have x go from 0 to a. set that volume to 1/4 the volume without the hole & solve for a .
     
  7. Feb 26, 2013 #6
    I'm getting two answers..2 and 2(21/2)
    and neither are correct.

    Should have included this, sorry. My integrand here has been changed to (x)(2-x2/2). When my bounds are [0,a] I get 2pi(a2 - a4/8). When my bounds are [0,2] I get the volume as 4pi. If I set the first part equal to 1/4 of the total volume I get 2pi(a2 -a4/8) = pi and I solve it from there to get that a = 21/2
     
    Last edited: Feb 26, 2013
  8. Feb 26, 2013 #7

    haruspex

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    Right
    Wrong.
     
  9. Feb 26, 2013 #8
    thanks, got it now.
     
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