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Integral help

  1. Oct 4, 2008 #1
    [tex]\int^{1.5}_{0}\int^{1.5}_{0}\int^{1.5}_{0}\frac{1}{x^2+y^2+z^2}dxdydz[/tex]

    I tried converting this to spherical and only integrating over a quarter of the octant but with no luck.

    Can someone please point me in the right direction.

    Thanks!
     
  2. jcsd
  3. Oct 4, 2008 #2

    gabbagabbahey

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    Do you know what
    [tex]\int_0^{1.5} \frac{dx}{x^2+a^2}[/tex]

    where [itex]a[/itex] is a constant is? If so, just integrate over x first holding y and z constant (i.e.[itex]a^2=y^2+z^2[/itex]) Then integrate over y, holding z constant. Lastly, integrate over z. Assuming that z,x,y have no functional relationship, and you limits of integration are correct, this is perfectly valid.
     
  4. Oct 4, 2008 #3
    Yes, but then I get to

    [tex]\int^{1.5}_{0}\int^{1.5}_{0}\frac{ArcTan(\frac{1.5}{\sqrt{y^2+z^2}})}{\sqrt{y^2+z^2}}dydz[/tex]

    and get stuck.
     
    Last edited: Oct 4, 2008
  5. Oct 4, 2008 #4

    gabbagabbahey

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    Hmmm... yes, Mathematica doesn't even want to do this integration. Is this really the original question?
     
  6. Oct 4, 2008 #5
    no, it was to find the time averaged power inside a box produced by light bulb in the center.
    This is a by product of assuming spherical wave propagation from the center of the box.
     
  7. Oct 4, 2008 #6

    gabbagabbahey

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    The cube is of side 1.5 or 3?
     
  8. Oct 4, 2008 #7
    It's a cube with side length 3, I was just considering one octant.
     
  9. Oct 4, 2008 #8

    gabbagabbahey

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    I see; and the power is proportional to [itex]\frac{1}{r^2}[/itex]. The only thing I can think of that might help is to use the fact that

    [tex]\vec{\nabla} \cdot \left( \frac{\hat{r}}{r} \right)=\frac{1}{r^2}[/tex]

    to turn the volume integral into a surface integral via the divergence theorem. Then compute the flux of [itex]\vec{v} \equiv \frac{\hat{r}}{r} [/itex] Through each face of the cube.
     
    Last edited: Oct 4, 2008
  10. Oct 4, 2008 #9
    Interesting...that's sounds like something worth pursuing, thanks for the help!
     
  11. Oct 4, 2008 #10

    gabbagabbahey

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    In fact, due to symmetry the flux through each face will be the same, so you only have to compute it for one face and then multiply by 6.
     
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