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Math Help - Cylindrical Shell Volume, why am I getting the wrong answer?

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    s3a
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    Cylindrical Shell Volume, why am I getting the wrong answer?

    My work is attached and the correct final answer is 250pi/3.

    Any help would be greatly appreciated!
    Thanks in advance!
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    Quote Originally Posted by s3a View Post
    My work is attached and the correct final answer is 250pi/3.

    Any help would be greatly appreciated!
    Thanks in advance!
    I think you left out the radius of the shell. Just multiply each of your integrands by y.
    Last edited by ione; February 24th 2010 at 09:43 AM.
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    Quote Originally Posted by s3a View Post
    My work is attached and the correct final answer is 250pi/3.

    Any help would be greatly appreciated!
    Thanks in advance!
    Hi s3a,

    I think you want to solve this using shells rather than washers.

    You haven't correctly set up the surface areas of the cylinders.

    For the cylinders whose heights go from one end of the curve to the other,
    corresponding to y=0 to y=4,

    r=y,\ h=2x=\sqrt{y}

    Then the vor for that region is 2{\pi}\int_{0}^4y\sqrt{y}dy=2{\pi}\int_{0}^4y^{\fr  ac{3}{2}}dy

    This works out to be about 25.6{\pi}


    For the upper region, where the cylinder height stretches between the line and curve

    r=y,\ h=x_{line}-x_{curve}

    x for the curve is negative here always.

    x_{curve}=-\frac{\sqrt{y}}{2}

    x_{line}=\frac{6-y}{2}

    h=\frac{6-y}{2}-\left(-\frac{\sqrt{y}}{2}\right)

    The vor of this region is 2{\pi}\int_{4}^9y\left(\frac{6-y}{2}+\frac{\sqrt{y}}{2}\right)dy={\pi}\int_{4}^9\  left(6y-y^2+y^{\frac{3}{2}}\right)dy

    Evaluating this yields about 57.73{\pi}

    The sum of the 2 volumes of revolution is (57.73+25.6){\pi}=83.33{\pi}
    Last edited by Archie Meade; February 23rd 2010 at 03:47 PM. Reason: small typo
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