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Math Help - Washer method

  1. #1
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    Washer method

    Find the volume generated by revolving the area bounded by x=y^2 and x=4 about the line y=2.

    Is it the integral from 0 to 4 of pi(4-(2-sqrt x)^2)?

    Sorry I don't know how to use latex yet.
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  2. #2
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    washer method ...

    R = 2 + \sqrt{x}

    r = 2 - \sqrt{x}

    R^2 - r^2 = (4 + 4\sqrt{x} + x) - (4 - 4\sqrt{x} + x) = 8\sqrt{x}

    V = \pi \int_0^4 8\sqrt{x} \, dx
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  3. #3
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    For shells, see if you get the same solution.

    2{\pi}\int_{0}^{4}(y-2)y^{2}dy
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  4. #4
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    Quote Originally Posted by skeeter View Post
    washer method ...

    R = 2 + \sqrt{x}

    r = 2 - \sqrt{x}

    R^2 - r^2 = (4 + 4\sqrt{x} + x) - (4 - 4\sqrt{x} + x) = 8\sqrt{x}

    V = \pi \int_0^4 8\sqrt{x} \, dx
    yeah i forgot that the graph is not bounded by y=0 and -sqrt(x) needs to be considered as well. Thanks for your help. appreciate it.

    And yeah I get the same answer with the shell method - 128pi/3
    Last edited by elitespart; November 19th 2008 at 04:23 PM.
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