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Thread: Volume by revolution

  1. February 3rd 2009, 03:31 PM #1
    silencecloak
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    Volume by revolution

    y=x
    y=0
    x=5
    x=6

    Rotate the area bounded by the given curves around x=1

    The easy numbers are throwing me off I think. My outside radius would just be 6-1 and my inner radius would just be 5 -1

    Correct?

    I have my integral set up as \int_0^6 (5^2-4^2)

    Did i do this problem correctly? Thanks
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  2. February 3rd 2009, 03:37 PM #2
    skeeter
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    method of cylindrical shells.

    rotation radius ... (x-1)

    shell height ... x

    limits ... x = 5 to x = 6

    V = 2\pi \int_5^6 (x-1)x \, dx
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