Integrating unit vectors on the surface of the unit sphere How can I calculate below integrals:
$$\int \hat{r}_i\ \hat{r}_j\ \hat{r}_k\ \hat{r}_l \ d\Omega$$
in which $d\Omega =\sin(\theta)
 \ d \theta\ d\phi $ is the surface element of the sphere and $\hat{r}_i$ indicates $i$'th component of unit vector r. For ex. $\hat{r}_1= \sin \theta \ \cos \phi$
Could anyone help me?
 A: Let me show you the universal trick in computing any integral of the form 
$$
I_{k_1...k_n}= \int d \Omega \, \hat r_{k_1}...\hat r_{k_n}.
$$
Just notice that (without being too formal) $I_{k_1...k_n}$ can be obtained by taking derivatives $\frac{\partial}{\partial j}$ of 
$$Z(j):=\int d \Omega \, e^{i j \cdot \hat r },$$
 and then setting $j=0$: $$
I_{k_1...k_n} = \frac{1}{i^{k_n}} \frac{\partial^{{k_n}}}{\partial j_{k_1} ... \partial j_{k_n}} Z(j) \vert_{j=0},
$$
..but we can easily compute the "partition function" $Z$! For three dimensions $r \in \mathbb{R}^3$ we have:$$
Z(j)= 4\pi \frac{sin j }{j}= 4\pi \left(1-\frac{j\cdot j}{3!}+ \frac{(j\cdot j)^2}{5!}+ \cdots\right),$$
where $j\cdot j$ is the euclidean scalar product.
A: Clearly the surface integral vanishes if it has a factor of any component to an odd power. We can always write the surface integral in spherical coordinates, aligning the $z$ axis with one of the indices appearing to the $4$-th or $2$-nd power. 
We can also make the substitution $u = \cos \theta$ where $du = -\sin \theta\, d\theta$.
Thus when $i=j=k=\ell$ the integral is 
$$
-\int_1^{-1}2\pi u^4 du=\frac45\pi
$$
(the $2\pi$ comes from thje integral over $\phi$.) 
And when $i=j$ and $k=\ell\neq i$ (and all other pairings of the like nature) the integral is
$$
-\int_{\phi = 0}^{2\pi}\cos^2\phi\int_1^{-1}u^2(1-u^2)  du
$$
where we have used $sin^2 = 1-\cos^2$ to turn the $\sin^2\theta$ into $(1-u^2)$.
The $d\phi$ integral is easy if you remember that over a full cycle, the average value of $\cos^2\phi$ is $\frac12$. We are left with
$$
\pi \int_{-1}^1(u^2-u^4)\,du=\pi\left( \frac23-\frac25 \right)=\frac2{15}\pi
$$
Finally, we need to turn these into a unified expression involving the indices $i,j,k,\ell$, which is allowed to used the Kroenecker delta $\delta_{mn} = 1$ if $m=n$ and $0$ otherwise.
The integral must then be of the form
$$
\int \hat{r}_i\hat{r}_j\hat{r}_k\hat{r}_\ell= \frac2{15} \pi
(\delta_{ij}\delta_{k\ell}+\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk})
+ \lambda \delta_{ij}\delta_{ik}\delta_{i\ell}
$$
where $\lambda$ is chosen to make the all-indices-equal answer come out to $\frac45\pi$. When all four indices are equal the first term becomes $\frac6{15}\pi = \frac25 \pi$ so $\lambda$ must be $\frac25$. The answer is
$$
\int \hat{r}_i\hat{r}_j\hat{r}_k\hat{r}_\ell= \frac2{15} \pi
(\delta_{ij}\delta_{k\ell}+\delta_{ik}\delta_{j\ell}+\delta_{i\ell}\delta_{jk}
+ 3\, \delta_{ij}\delta_{ik}\delta_{i\ell})
$$
A: The first integral always vanishes. To see this, write 
$$I=\int_{\mathbb S^2} r_i r_j r_k\, d\Omega.$$
Suppose that $r_i$ appears with power $1$ or $3$ (if it appears with power $2$, then choose the other one). Then the change of variable $r_i\mapsto -r_i, r_j\mapsto r_j$ for $j\ne i$ leaves $d\Omega$ invariant and shows that 
$$I=-I.$$
For the same reason, the second integral vanishes if there is a factor of $r_i$ appearing with power $1$ or $3$. The only nonvanishing integrals are therefore 
$$J_1=\int_{\mathbb S^2} r_i^2 r_j^2\, d\Omega, \quad J_2=\int_{\mathbb S^2} r_i^4\, d\Omega.$$ 
Here you can assume without loss of generality that $r_i=z=\cos \theta, r_j=y=\sin\theta\cos \phi$. The integrals become
$$
J_1=\int_0^\pi (\cos\theta)^2(\sin\theta)^3\, d\theta \int_0^{2\pi} (\cos\phi)^2\, d\phi, \quad J_2=2\pi\int_0^\pi (\cos \theta)^4 \sin \theta\, d\theta,$$ 
and they can be computed directly.
