Math (Calculus)

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis.

y=4-x^2, y=0

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asked by Sherman
  1. The volume of a shell of radius r, height h and thickness dy (because revolving around the y-axis) is

    2πrh dx

    Now, our plane region is symmetric about the y-axis, so revolving all of it it is the same as revolving only the part for x>=0. o, we'll take the volume to be

    ∫[0,2] 2πrh dx
    where r=x and h=y
    = 2π∫[0,2] x(4-x^2) dx
    = 2π(2x^2 - 1/4 x^4) [0,2]
    = 8π

    You can check using shells of thickness dy. The volume is then

    ∫[0,4] πr^2 dy
    where r=x
    = π∫[0,4] (4-y) dy
    = π(4y - 1/2 y^2) [0,4]
    = 8π

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    posted by Steve

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