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Math Help - Double Integral Question

  1. #1
    DCU
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    Double Integral Question



    Will the answer to this be 1/6?
    I got limts as

    upper x = 1 lower x = sqrt(1-y^2)
    upper y = 1 lower y = 0

    This right?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by DCU View Post


    Will the answer to this be 1/6?
    I got limts as

    upper x = 1 lower x = sqrt(1-y^2)
    upper y = 1 lower y = 0

    This right?
    I got a different answer. It say to switch to polar so

     <br />
\int_0^{\pi}\int_0^1 (r \cos \theta + r \sin \theta + 1)r dr d \theta = \frac{2}{3} + \frac{\pi}{2}<br />
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  3. #3
    DCU
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    Quote Originally Posted by Danny View Post
    I got a different answer. It say to switch to polar so

     <br />
\int_0^{\pi}\int_0^1 (r \cos \theta + r \sin \theta + 1)r dr d \theta = \frac{2}{3} + \frac{\pi}{2}<br />
    I must have read it wrong. Thanks
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