Integral help

zedmed

$$\int^{1.5}_{0}\int^{1.5}_{0}\int^{1.5}_{0}\frac{1}{x^2+y^2+z^2}dxdydz$$

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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gabbagabbahey

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Do you know what
$$\int_0^{1.5} \frac{dx}{x^2+a^2}$$

where $a$ is a constant is? If so, just integrate over x first holding y and z constant (i.e.$a^2=y^2+z^2$) 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.

zedmed

Yes, but then I get to

$$\int^{1.5}_{0}\int^{1.5}_{0}\frac{ArcTan(\frac{1.5}{\sqrt{y^2+z^2}})}{\sqrt{y^2+z^2}}dydz$$

and get stuck.

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gabbagabbahey

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Hmmm... yes, Mathematica doesn't even want to do this integration. Is this really the original question?

zedmed

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.

gabbagabbahey

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The cube is of side 1.5 or 3?

zedmed

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

gabbagabbahey

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

$$\vec{\nabla} \cdot \left( \frac{\hat{r}}{r} \right)=\frac{1}{r^2}$$

to turn the volume integral into a surface integral via the divergence theorem. Then compute the flux of $\vec{v} \equiv \frac{\hat{r}}{r}$ Through each face of the cube.

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zedmed

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

gabbagabbahey

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Gold Member
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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