- #1

TheGreatDeadOne

- 11

- 0

- Homework Statement:
- Let a sphere of radius r_0 be centered at the origin, and r′ the position vector of a point p′ within the sphere or under its surface S. Let the position vector r be an arbitrary fixed point P.

- Relevant Equations:
- Divergence theorem (it's not really necessary, but I'm trying to solve it that way)

The integral that I have to solve is as follows:

[tex]\oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}[/tex], integrating with respect to r'

Then I apply the divergence theorem, resulting in:

[tex]\iiint \limits _{v} \nabla \cdot \frac{1}{|r-r'|}dv' = \int_{0}^{r_0}\int_{0}^{\pi}\int_{0}^{2\pi} \frac{(r-r')}{|r-r'|^3}r'^2\sin{\theta}d\theta d\phi dr'[/tex]

Integrating θ and φ we have:

[tex]4\pi \int_{0}^{r_0} \frac{(r-r')}{|r-r'|^3}r'^2 dr'[/tex]

And in that part I got stuck. I didn't try to solve it by integrating directly, because I'm still a bit confused. I know the result is:

[tex]\frac{4\pi r^{2}_{0}}{r} \quad\text{for} \quad r\geq r_0 \quad and \quad 4\pi r^{2}_{0}\quad for \quad r\leq r_0[/tex]

[tex]\oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}[/tex], integrating with respect to r'

Then I apply the divergence theorem, resulting in:

[tex]\iiint \limits _{v} \nabla \cdot \frac{1}{|r-r'|}dv' = \int_{0}^{r_0}\int_{0}^{\pi}\int_{0}^{2\pi} \frac{(r-r')}{|r-r'|^3}r'^2\sin{\theta}d\theta d\phi dr'[/tex]

Integrating θ and φ we have:

[tex]4\pi \int_{0}^{r_0} \frac{(r-r')}{|r-r'|^3}r'^2 dr'[/tex]

And in that part I got stuck. I didn't try to solve it by integrating directly, because I'm still a bit confused. I know the result is:

[tex]\frac{4\pi r^{2}_{0}}{r} \quad\text{for} \quad r\geq r_0 \quad and \quad 4\pi r^{2}_{0}\quad for \quad r\leq r_0[/tex]