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Thread: Volume by revolution

  1. #1
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    Volume by revolution

    $\displaystyle y=x$
    $\displaystyle y=0$
    $\displaystyle x=5$
    $\displaystyle x=6$

    Rotate the area bounded by the given curves around $\displaystyle x=1$

    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 $\displaystyle \int_0^6 (5^2-4^2)$

    Did i do this problem correctly? Thanks
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  2. #2
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    method of cylindrical shells.

    rotation radius ... $\displaystyle (x-1)$

    shell height ... $\displaystyle x$

    limits ... $\displaystyle x = 5$ to $\displaystyle x = 6$

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