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Cylindrical Shell Method

  1. Sep 3, 2006 #1
    Please help! How do I do this problem?

    Using the method of cylindrical shells, find the volume generated by rotating the region the region bounded by the given curves about the specified axis.

    y=(x-1)^(1/2), y=0, x=5; about y = 3

    Please tell me how to set up the integral! Any help is MUCH appreciated.

    So far I have Integral from 0 to 2 of (3-(y^2+1))*y dy I know that isn't right, because I am not getting the right answer! The book says that it is 24pi.

    Thank you.
  2. jcsd
  3. Sep 3, 2006 #2


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    First, draw a picture. Then cut the region into horizontal slabs of height dy (or [itex]\Delta y[/itex] if you prefer). Imagine a slab at height y being rotated around the line y=3. What would be the volume the obtained cylinder?
    Next, add the contributions from all cylinders, i.e. integrate.
  4. Sep 4, 2006 #3
    I think that using vertical cylindrical shells would be most effective for this problem.

    I would use vertical cylindrical shell whose outer radius r2 is [tex]3+\sqrt{x-1}[/tex] while its inner radius r1 is 3 and whose height is dx.

    Now the infinitesimal volume of any arbitrary shell would be

    [tex]dV = \pi\cdot dx\cdot \big[ (3+\sqrt{x-1})^2 - 3^2) \big][/tex]

    Now since x would have to go from 1 to 5 to cover the required region, we just integrate the above expression from 1 to 5 and we get:

    [tex]V = \pi\cdot\int_0^5 \big[ (3+\sqrt{x-1})^2 - 3^2) \big] dx[/tex]

    This should give you the right answer.
    Last edited: Sep 4, 2006
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