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Integrating in Spherical Co-Ordinates.

  1. May 9, 2005 #1
    I have the following Integral

    [tex] \int ^1 _0 [/tex] [tex] \int _0 ^\sqrt{1-x^2} [/tex] [tex] \int _0 ^\sqrt{1-x^2-y^2} [/tex] [tex] \frac{1}{1+(x^2)+(y^2)+(z^2)} dzdydx [/tex]

    (With the limits working properly!)

    Converted to spherical Cor-ordinates, I have

    [tex] \int ^\frac{\pi}{2} _0 [/tex] [tex] \int _0 ^\frac{\pi}{2} [/tex] [tex] \int _0 ^1 [/tex] [tex] \frac{1}{1+\rho} \rho^2 sin(\phi) d\rho dr d\phi [/tex]

    I've converted the function, but how would I start integrating?
     
    Last edited: May 9, 2005
  2. jcsd
  3. May 9, 2005 #2

    dextercioby

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    Alright.Transform it in spherical coordinates.But u need the limits...

    Daniel.
     
  4. May 9, 2005 #3
    I have the limits.... it's just that I can't get them right on Latex (stupid coding! I'm getting it right, it's just not displaying it that way!)

    Hang on.....

    EDIT: Problem above now has working limits, and my original question as intended.
     
    Last edited: May 9, 2005
  5. May 9, 2005 #4
    A substitution MIGHT work, but i would probably go for integration by parts. Remember sin(phi) is constant for the first integral.
     
  6. May 9, 2005 #5

    dextercioby

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    [tex] \frac{r^{2}}{1+r}=r-1+\frac{1}{1+r} [/tex]

    is all u need.

    Daniel.
     
  7. May 9, 2005 #6

    arildno

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    Your integrand in spherical coordinates should be: [tex]\frac{\rho^{2}\sin\phi}{1+\rho^{2}}[/tex]
     
  8. May 9, 2005 #7
    I've got this thanks. Thanks to a form of integration in the back of my book.
     
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