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Math Help - volume problem - cylindrical shell method

  1. #1
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    volume problem - cylindrical shell method

    "The region enclosed by the curves y=x^{3} and y=\sqrt{x} is roated about the line x=1. Fint the volume of the resulting solid" -James Stewart

    answer: \frac{13\pi}{30}
    what I did:
    difference of shells formula is 2\pi \int x(f(x)-g(x))
    so in this problem f(x)=\sqrt{x} and g(x)=x^3. The intersections are x=0,1. I get confused because it's not being rotated around the y-axis but x=1
    I tried V= 2 \pi \int_0^1 x((1-\sqrt{x})-(1-x^{3}) but I end up with a negative number.
    Last edited by superdude; July 20th 2009 at 08:36 PM. Reason: missing closing bracket
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  2. #2
    MHF Contributor Calculus26's Avatar
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    The x term represents the distance between an arbitrary x and the axis of rotation.

    If the solid is rotated about the yaxis the x term is just x.

    In your example the distance between x and the axis of rotation x = 1 is

    1-x. so replace x with 1-x--- you obtain (1-x)*(sqrt(x)-x^3) for the integrand.

    A more general form of the shell method is 2pint{D(x)*(f-g)dx}

    Where D(x) is as mentioned the distance between an arbitrary x and the axis of rotation
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