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Volumes: The Disk Method

  1. Nov 9, 2008 #1
    Volumes: The Disk Method [Resolved]

    1. If the area bounded by the parabola y = H - (H/R[tex]^{2}[/tex])x[tex]^{2}[/tex] and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolition with height H and radius of base R. Show its volume is half the volume of the circumscribing cylinder

    Okay so the thickness of the disk is dy and the area is [tex]\pi[/tex]x[tex]^{2}[/tex]. How do I find the limits of intergration and put x in terms of H and R? Thanks.
    Last edited: Nov 9, 2008
  2. jcsd
  3. Nov 9, 2008 #2


    Staff: Mentor

    Your disks run from y = 0 to y = H.
    The incremental volume of each disk is [tex]\pi x^2 \Delta y[/tex]. For any point (x, y) on the parabola, you have a relationship between x and y, so use this relationship to replace x^2.
  4. Nov 9, 2008 #3
    So x^2 = y-H(-R^2/H) and then run that from 0 to H?
  5. Nov 9, 2008 #4
  6. Nov 9, 2008 #5


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    sounds good, what do you get when you integrate that?
  7. Nov 9, 2008 #6
    I feel like an idiot but I'm having trouble integrating this. Can someone get me started on it? Thanks.
  8. Nov 9, 2008 #7


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    [tex]\int_0^H \pi x^2 dy= \int_0^H \pi (y-H)(\frac{-R^2}{H}) dy =\frac{\pi R^2}{H} \int_0^H (H-y)dy[/tex]

    ....surely, you know how to integrate this?
  9. Nov 9, 2008 #8
    wooow. there's something wrong with me. So I ended up with the volume of a cylinder over 2. Which is exactly what I needed. Thank you very much.
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