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Volume of a solid of revolution

  1. Jul 7, 2010 #1
    Can anyone confirm if I have done the following work correctly

    Find the volume of a solid of revolution obtained by rotating about the y axis the region bounded by y = the fifth root of x and 2x^2 - 3x + 2.

    By drawing the graph, I figured out that I need to use the method of cylindrcal shells given by v = integral from 0.619 to 1 of A(x) dx.

    Where A(x) = 2pi(radius)(height)

    The intersection points of the equations are approximately x = 1 and x = 0.619

    Radius is equal to x.

    Height is equal to the difference in the two equations

    i.e. (x^(1/5) - 2x^2 + 3x - 2)

    Thus we have 2pi(radius)(height)

    =2pi*(x)*(x^(1/5) - 2x^2 + 3x - 2)

    = 2pi*(x^(6/5) - 2x^3 + 3x^2 - 2x)

    Now I will integrate this between 0.619 and 1

    = (5/11)x^(11/5) - (1/2)x^4 + x^3 - x^2 ¦ 0.619 to 1
    which gives me 2pi*(-0.05 + 0.1345)
    =0.169pi approximately.

    Can anyone confirm that I have done this correctly
  2. jcsd
  3. Jul 7, 2010 #2


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    Your setup looks OK. Maple gives V := .03140665150*Pi.
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