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Cylindrical shells

  1. Jul 19, 2005 #1
    Hi I'm stuck on an integration problem where I need to use the method of cylindrical shells to calculate a volume.

    Q. Using the method of cylindrical shells, find the volume generated when the area bounded by the curve y = x^2 - 3 and the line y = 2x is revolved about the line x = 7.

    The main thing that I'm having trouble with is setting up the integral. I think it's due to a limited conceptual understanding of what's going on with questions of this type. I started by drawing a quick sketch and finding the x values of intersection, x = +/- sqrt(3).

    I drew a cylinder, with the vertical axis x = 7 going through its centre, around the region.

    Volume(cylinder) = (circumference)(height)(thickness) = CHT

    [tex]
    C = 2\pi r = 2\pi \left( {7 - x} \right)
    [/tex]

    [tex]
    H = 2x - \left( {x^2 - 3} \right) = - x^2 + 2x + 3
    [/tex]

    T = dx

    [tex]
    V = 2\pi \int\limits_{ - \sqrt 3 }^{\sqrt 3 } {\left( {7 - x} \right)\left( { - x^2 + 3x + 3} \right)dx}
    [/tex]

    That's what I came up with. I can't check to see if my answer is correct because I don't have the solution to the question. Any help with this question would be good thanks.
     
  2. jcsd
  3. Jul 19, 2005 #2

    HallsofIvy

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    Well, the first thing you've done wrong is miscalculate the x values of the intersection of the curves. They are not [tex]\plusminus\sqrt{2}[/tex]!
    Show how you got that value and we may be able to point out your mistake.
    Other than that, it looks like you have the right integral.
     
  4. Jul 19, 2005 #3

    saltydog

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    So what'd ya get Benny? To find the intersection of two curves f(x) and g(x), just set them equal to each other right? So set those two equal, solve for x and you have the limits of integration. I made a plot.
     

    Attached Files:

  5. Jul 20, 2005 #4
    Hmm...x^2 - 3 = 2x => (x-3)(x+1) = 0 => x = 3 or -1. I don't know how I got +/- sqrt(3) the first time around. Thanks for pointing that out.

    So continuing from before(the quadratic should have a 2x term instead of 3x in my previous post).

    [tex]
    V = 2\pi \int\limits_{ - 1}^3 {\left( {7 - x} \right)} \left( { - x^2 + 2x + 3} \right)dx
    [/tex]

    [tex]
    = 2\pi \int\limits_{ - 1}^3 {\left( {x - 7} \right)\left( {x^2 - 2x - 3} \right)dx}
    [/tex]

    [tex]
    = 2\pi \int\limits_{ - 1}^3 {\left( {x^3 - 9x^2 + 11x + 21} \right)dx}
    [/tex]

    [tex]
    = 2\pi \left[ {\frac{{x^4 }}{4} - 3x^3 + \frac{{11x^2 }}{2} + 21x} \right]_{ - 1}^3
    [/tex]

    [tex]
    = 2\pi \left( {\left( {\frac{{81}}{4} - 81 + \frac{{99}}{2} + 63} \right) - \left( {\frac{1}{4} + 3 + \frac{{11}}{2} - 21} \right)} \right)
    [/tex]

    [tex]
    \mathop = \limits^{calculator} 128\pi
    [/tex]

    Edit: BTW saltydog, which program do you use to draw those sketch graphs?
     
  6. Jul 20, 2005 #5

    saltydog

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    That's what I got. I use Mathematica. Try and check it out. Your school should have it. You know, Mathematica is also good to back-substitute solutions into ODEs to verify the solutions. Good for lots of other things too.:smile:
     
  7. Jul 20, 2005 #6

    OlderDan

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    Some Mathematica functions are avilable online at

    http://www.quickmath.com/

    Click on equations and advanced equation plotting page
     
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