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Math Help - integral

  1. #1
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    integral

     \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} \int_{-\sqrt{4-x^{2}-y^{2}}}^{\sqrt{4-x^{2}-y^{2}}} y^{2} \sqrt{x^{2}+y^{2}+z^{2}} \ dz \ dx \ dy

    I got it down to  \int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{2} \rho^{5} \sin^{3}\phi \ \sin^{2} \theta \ d \rho \ d \theta \ d \phi

    And from here just separate it out and integrate.

    Is this correct?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by shilz222 View Post
     \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} \int_{-\sqrt{4-x^{2}-y^{2}}}^{\sqrt{4-x^{2}-y^{2}}} y^{2} \sqrt{x^{2}+y^{2}+z^{2}} \ dz \ dx \ dy

    I got it down to  \int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{2} \rho^{5} \sin^{3}\phi \ \sin^{2} \theta \ d \rho \ d \theta \ d \phi

    And from here just separate it out and integrate.

    Is this correct?

    im not sure but i think, you'll just need to integrate one and consider the others as constants and evaluate, then integrate again and so on.. its just the same way you did in the double integrals..
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  3. #3
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    Quote Originally Posted by shilz222 View Post
    Is this correct?
    Yes. Just integrate part by part. And the other note is that it can be done in any order because now \theta - \phi - \rho determine a rectangular prism and you can integrate in any way (any order) over it.
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