# Formidable vector integral

Suppose $\vec{M}, \vec{P}$ are arbitrary, constant vectors, and $\hat{r}$ is the (unit) position vector in spherical polar coordinates.

I need to integrate the vector function $\frac{1}{r^6}[\hat{r} \times ((\vec{P}\cdot \hat{r})\vec{M} - ((\vec{M} \cdot \hat{r})\vec{P})]$ over the entire exterior of the sphere of radius R centered at the origin of coordinates. In other words, I need to compute:

$$\int_{\phi = 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r = R}^{\infty} \frac{1}{r^4}[\hat{r} \times ((\vec{P}\cdot \hat{r})\vec{M} - ((\vec{M} \cdot \hat{r})\vec{P})] sin\theta dr d\theta d \phi$$

I'm looking for a cute and clever way to do this, instead of the straightforward and tedious method. Any ideas or hints?

fzero
Homework Helper
Gold Member
The slickest way to do this is to write

$$[r \times ((\vec{P}\cdot r)\vec{M} - ((\vec{M} \cdot r)\vec{P})]_i = \sum_{jklmn} \epsilon_{ijk} r_j r_l P_m M_n ( \delta_{lm}\delta_{kn} - \delta_{ln}\delta_{km} )$$

and write the $$r_i$$ in terms of spherical harmonics:

$$\frac{r_1}{r} = - \sqrt{\frac{2\pi}{3}} \left( Y_1^1 + Y_1^{-1} \right),$$

$$\frac{r_2}{r} = i \sqrt{\frac{2\pi}{3}} \left( Y_1^1 - Y_1^{-1} \right),$$

$$\frac{r_3}{r} = 2 \sqrt{\frac{\pi}{3}} Y_1^0 .$$

The angular integration is done using the orthogonality relations and gives you a matrix $$Q_{jl}$$ that you then have to sum over. Whether it saves that much work over brute force is to be determined.