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Math Help - volume of revolution

  1. #1
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    volume of revolution

    hello guys, just wanted to see if i derived this question correctly.

    A bowl is formed by rotating about the y-axis the area contained between that part of the curve 2y=x^2 from x = 0 to x = 4, the line x=4 and the x-axis. calculate the maximium volume of water he bowl could hold.

    i have derived the formula

    Volume = \int^0 \pi (x^2/2)^2 dx

    the upper limit is 0 and the lower limit should be 4. i cant seem to get that in latex form..
    is this correct though
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by sigma1 View Post
    hello guys, just wanted to see if i derived this question correctly.

    A bowl is formed by rotating about the y-axis the area contained between that part of the curve 2y=x^2 from x = 0 to x = 4, the line x=4 and the x-axis. calculate the maximium volume of water he bowl could hold.

    i have derived the formula

    Volume = \int^0 \pi (x^2/2)^2 dx

    the upper limit is 0 and the lower limit should be 4. i cant seem to get that in latex form..
    is this correct though
    Well lets look at it like this. We have a parabola bounded by the vertical lines x=0, x=4 and the horizontal line y=0

    So this would appear to be best for the method of cylindrical shells.

    The interior of the bowl corresponds to the revolving region given by

    x^2/2<=y<=8

    The area element at position x has height

    8-x^2/2

    and generates the clindrical shell of volume

    dV = 2\pi x(8-x^2/2)

    Edit- Was it just my imagination or didn't I see a moderator post in this thread? They had something different then me but I don't see why what I've done would be wrong. Although it has been a while since I did cylindrical shells. Can you post again?
    Last edited by AllanCuz; April 11th 2010 at 04:18 PM.
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  3. #3
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    disks w/r to y ...

    V = \pi \int_0^8 2y \, dy
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