1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?