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Math Help - Volume of Rotation of a Quarter Circle

  1. #1
    Junior Member
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    Volume of Rotation of a Quarter Circle

    I am trying to find the volume of a quarter circle that is rotated about x=1. So the equation should look like this:
    <br />
\int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy

    This should simplify to:
    \int_0^{\frac {\pi} {2}} \pi(1-y^2)dy

    If I am doing this right thus far, then the integral is:
    \pi(y-\frac {y^3} {3}) from 0 to \frac {\pi} {2}

    And when I plug these values in I some how end up with \frac {\pi^2} {2} - \frac {\pi^4} {24}

    I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Dec 2008
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    South Coast of England
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    Hello Spudwad
    Quote Originally Posted by Spudwad View Post
    I am trying to find the volume of a quarter circle that is rotated about x=1. So the equation should look like this:
    <br />
\int_0^{\frac {\pi} {2}} \pi(\sqrt {1^2-y^2})^2dy

    This should simplify to:
    \int_0^{\frac {\pi} {2}} \pi(1-y^2)dy

    If I am doing this right thus far, then the integral is:
    \pi(y-\frac {y^3} {3}) from 0 to \frac {\pi} {2}

    And when I plug these values in I some how end up with \frac {\pi^2} {2} - \frac {\pi^4} {24}

    I feel like I am missing something in between here, or maybe just made an error plugging in the values, although I have tried it a few times and end up with the same answer. Is this right? Thanks for the help.
    Get ready to kick yourself:
    The limits of the integral are 0 to 1, not 0 to \pi/2.

    Grandad
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