Solving Poyntings Vector for a point charge

[HPRF]

Homework Statement

It is found that Poyntings vector gives

P = ExH = (mu₀q²a²sin²(theta)/6pi²cr²)r

This apparently leads to

Total Power = (mu₀q²a²/6pi²c)$$\int$$(sin²(theta)/r²)(2pir²sin(theta)d$$\theta$$)

What I am unsure of is where the

(2pir²sin(theta)d$$\theta$$)

appears from. Can anyone help?

 

[jdwood983]

[strike]The power is over the whole of the volume, and not just one component of the volume.[/strike] When you are coming from Cartesian coordinates to http://en.wikipedia.org/wiki/Spherical_coordinate_system#Cartesian_coordinates", we find

$$\int dV=\int_{-\infty}^\infty dx\int_{-\infty}^\infty dy\int_{-\infty}^\infty dz=\int_0^\infty r^2dr\int_0^\pi \sin\theta \,d\theta\int_0^{2\pi} d\phi$$

 

[gabbagabbahey]

The power is over the whole of the volume[/color], and not just one component of the volume.

Errr... not quite...Poynting's theorem tells us that the power crossing a surface $\mathcal{S}$ is given by

$$P(\textbf{r})=\int_{\mathcal{S}}\textbf{S}\cdot d\textbf{a}$$

(using $\textbf{S}$ to represent the Poynting vector and $d\textbf{a}$ to represent the infinitesimal vector area element of the surface)

To find the power radiated, one usually uses a spherical surface of radius $r$ (so that [itex]d\textbf{a}=r^2\sin\theta d\theta d\phi\mathbf{\hat{r}}[/tex])) and then takes the limit as $r\to\infty$ (since radiation escapes to infinity).[/itex]