Integrating Over an Octant: Help Needed

In summary, the conversation discusses converting a triple integral to spherical coordinates in order to find the time averaged power inside a cube produced by a light bulb in the center. It also suggests using the divergence theorem to turn the volume integral into a surface integral and computing the flux through each face of the cube.
  • #1
zedmed
5
0
[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!
 
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  • #2
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.
 
  • #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:
  • #4
Hmmm... yes, Mathematica doesn't even want to do this integration. Is this really the original question?
 
  • #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.
 
  • #6
The cube is of side 1.5 or 3?
 
  • #7
It's a cube with side length 3, I was just considering one octant.
 
  • #8
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.
 
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  • #9
Interesting...that's sounds like something worth pursuing, thanks for the help!
 
  • #10
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.
 

1. What is meant by "integrating over an octant"?

Integrating over an octant refers to the process of finding the total value of a function within a specific region or volume defined by an octant (one eighth of a three-dimensional coordinate system).

2. Why is integrating over an octant important in scientific research?

Integrating over an octant allows scientists to calculate important quantities such as volume, mass, or charge distribution in three-dimensional space. It is commonly used in fields such as physics, engineering, and mathematics to solve problems and make predictions.

3. How is integrating over an octant different from other integration techniques?

Integrating over an octant only considers a specific region of a three-dimensional space, while other integration techniques may cover larger or more complex regions. It also involves using specific formulas and limits based on the octant being integrated over.

4. What are some challenges that may arise when integrating over an octant?

One potential challenge is determining the correct limits of integration for a given octant. Another challenge may be dealing with functions that are not continuous or have discontinuities within the octant being integrated over.

5. Can integrating over an octant be applied to real-world problems?

Yes, integrating over an octant can be applied to various real-world problems such as calculating the mass of a 3D object, finding the electric field within a certain volume, or determining the volume of a chemical reaction vessel. It is a powerful tool for solving problems in physics, engineering, and other fields.

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