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Math Help - Nested integral in spherical coordinates

  1. #1
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    Nested integral in spherical coordinates

    I am suppose to evaluate this integral in spherical coordinates

    f(x,y,z) = 1/ (x^2 + y^2 + z^2)^(1/2) over the bottom half of the sphere of radius 5 centered at the origin.

    I believe I am to take the integral of 1/ roe (roe^2 sin phi) d roe d phi d theta
    so it is just roe times sin phi as the function but the bounds are hard for me

    theta of course is from 0 to two pi
    I know z goes from -5 to 0

    but how to turn this into bounds in spherical coordinates???? Frostking
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Frostking View Post
    I am suppose to evaluate this integral in spherical coordinates

    f(x,y,z) = 1/ (x^2 + y^2 + z^2)^(1/2) over the bottom half of the sphere of radius 5 centered at the origin.

    I believe I am to take the integral of 1/ roe (roe^2 sin phi) d roe d phi d theta
    so it is just roe times sin phi as the function but the bounds are hard for me

    theta of course is from 0 to two pi
    I know z goes from -5 to 0

    but how to turn this into bounds in spherical coordinates???? Frostking
    \rho is the new "radius". It goes from 0 to 5.

    \theta remains the same.

    But \phi goes from \frac{\pi}{2} to \pi if we are considering the sphere below the z=0 plane.

    Now, the function is the same as \frac{1}{\rho}.

    So you will be integrating \int_0^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\int_0^5\fr  ac{1}{\rho}\rho^2\sin\phi\,d\rho\,d\phi\d,\theta=\  int_0^{2\pi}\int_{\frac{\pi}{2}}^{\pi}\int_0^5\rho  \sin\phi\,d\rho\,d\phi\d,\theta

    Can you take it from here?
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  3. #3
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    nested integral in spherical terms

    Yes! Thanks very much for your effort. Frostking
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