How Do You Integrate Over an Octant for \(\frac{1}{x^2+y^2+z^2}\)?

[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!

 

[gabbagabbahey]

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.

 

[gabbagabbahey]

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]

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]

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.

 

[zedmed]

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

 

[gabbagabbahey]

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.