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Math Help - Triple integral

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face Triple integral

    Evaluate the following using cylindrical coordinates:
    \int\int\int_E(x^3+xy^2)dV
    where E is the solid that lies in the first octant beneath z=16-x^2-y^2
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Em Yeu Anh View Post
    Evaluate the following using cylindrical coordinates:
    \int\int\int_E(x^3+xy^2)dV
    where E is the solid that lies in the first octant beneath z=16-x^2-y^2
    Hi

    x^2+y^2 =16-z means that z varies from 0 to 16

    x^2+y^2 = r^2 = 16-z therefore for any value for z between 0 and 16, r varies from 0 to \sqrt{16-z}

    The first octant means that \theta varies from 0 to \frac{\pi}{2}

    Then x^3+xy^2 = r^3 \cos^3 \theta + r \cos \theta r^2 \sin^2 \theta = r^3 \cos \theta

    and dx dy is to be replaced by r dr d\theta

    \int\int\int_E(x^3+xy^2)dV = \int_{0}^{16} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{16-z}} r^4 \cos \theta dr d\theta dz
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  3. #3
    MHF Contributor
    Jester's Avatar
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    What I might do is

     <br />
\int_0^{\pi/2} \int_0^4 \int_0^{16-r^2} r^4 \cos \theta dz\,dr\, d \theta<br />

    it would avoid the \sqrt{\;\;\;}.
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