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Cylindrical Shells

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the method of cylindrical shells to find the volume generated
    by rotating the region bounded by the given curves about the
    specified axis. Sketch the region and a typical shell.

    y = 4x - x^2, y = 3; about x = 1


    2. Relevant equations

    ?

    3. The attempt at a solution

    int(2*pi*(1-x)*(3-4*x+x^2), x = 0 .. 1) = 11pi/6

    This is the only problem on the book that I can't really understand. The correct answer is 8pi/3 according on the odd answers on the back page.

    Increasing the upper limit to 2 gives the correct answer. But why? Isn't it I'll just integrate until 1 so I can get a y = 3. Why 2?
     
  2. jcsd
  3. Dec 17, 2009 #2

    Mark44

    Staff: Mentor

    Your limits of integration are wrong. The region over which you are integrating is [1, 3], not [0, 1].
    I get 8pi/3 with this integral:
    [tex]-2\pi \int_1^3 (x - 1)(x^2 - 4x + 3)dx[/tex]
     
  4. Dec 17, 2009 #3
    I think the mistake you are making is in the formula that you are integrating. It seems like you were thinking washers while using the shell equation.
    The general equation to use for the shell method is

    2[tex]\pi[/tex][tex]\int[/tex] R dx(dx can change depending on which variable you are integrating with respect to, but in this case it would be dx).

    Where R is the distance from the axis of rotation. In this case, the R isn't 3-f(x), just f(x).
     
  5. Dec 17, 2009 #4

    Mark44

    Staff: Mentor

    Sarah's integrand was correct. Her only problem was she had the wrong limits of integration.
    No, this is not correct. The volume of a cylindrical shell is 2pi * radius * height * thickness. Your formula completely omits the height of the shell.
     
  6. Dec 17, 2009 #5
    why would the limits of integration be from 1 to 3 if you are integrating with respect to the x axis?
     
  7. Dec 17, 2009 #6
    nevermind, i've got it! sorry for my faulty answer and thank you for correcting my mistake.
     
  8. Dec 17, 2009 #7

    Mark44

    Staff: Mentor

    No problem. We all make mistakes at least once in a while...
     
  9. Dec 17, 2009 #8
    oh... now i noticed my problem.

    4x - x^2 = 3
    0 = x^2 - 4x + 3
    0 = (x-3)(x-1)

    x = 1 and 3

    Tiredness really made my brain crazy.
     
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