- #1

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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!

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- Thread starter zedmed
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- #1

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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

gabbagabbahey

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[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.

- #3

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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.

[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.

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- #4

gabbagabbahey

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- #5

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This is a by product of assuming spherical wave propagation from the center of the box.

- #6

gabbagabbahey

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The cube is of side 1.5 or 3?

- #7

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It's a cube with side length 3, I was just considering one octant.

- #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.

[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.

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- #9

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Interesting...that's sounds like something worth pursuing, thanks for the help!

- #10

gabbagabbahey

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