Surface integral of normal components summations on a sphere Whilst studying a book on fluid dynamics I came across a curious footnote comment which is essential in another derivation. The footnote states the following two identities:
$$
\frac{1}{|S|}\oint_S\! n_in_j\, dA = \frac{1}{3} \delta_{ij}\\
\frac{1}{|S|}\oint_S\! n_i n_j n_k n_l\, dA =
\frac{1}{15} (\delta_{ij} \delta_{lm} + \delta_{ik}\delta_{jm} + \delta_{im}\delta_{jl} )
$$
where $S$ is a unit sphere, $n_i$ is a normal and $|\cdot|$ is the surface area. It looks like something pretty standard from vector calculus, but I cannot find it right now. Would anyone like to point me to a proof of these? Is there like a definitve compedium with proofs of these type of identities?
 A: Here's a unifying principle:$\,\,$ Integrating the dyadic product of radial vectors over a sphere will result in an isotropic tensor.
So assume 
$$\oint \hat{n}\,\hat{n}\,dA = \beta I$$
for some undetermined scale factor $\beta$.
The scale factor can be determined by contraction
$$\eqalign{
  I:I &= \delta_{ij}\,\delta_{ij} = 3 \cr
  3\,\beta &= \oint \hat{n}\cdot\hat{n}\,dA \cr &= \oint dA = 4\pi R^2 \cr
  \beta &= \frac{4\pi}{3} R^2 \cr
\cr
}$$
Similarly
$$\oint \hat{n}\,\hat{n}\,\hat{n}\,\hat{n}\,dA = \beta\,{\mathbb Y}$$
where ${\mathbb Y}$ is a 4th order isotropic tensor. There are 3 such tensors, so in general ${\mathbb Y}$ is the linear combination
$${\mathbb Y}_{ijkl} = \beta_1\,\delta_{ij}\,\delta_{kl} + \beta_2\,\delta_{ik}\,\delta_{jl} + \beta_3\,\delta_{il}\,\delta_{jk} $$
We obviously need the result which is symmetric over every index, i.e. $$\beta_1=\beta_2=\beta_3=1$$
The contraction  $\big\{I:{\mathbb Y}:I\big\}$ yields 
$$\eqalign{
\delta_{ij}\,(\delta_{ij}\,\delta_{kl} + \delta_{ik}\,\delta_{jl} + \delta_{il}\,\delta_{jk})\,\delta_{kl} 
  &= (3\,\delta_{kl} + \delta_{kl} + \delta_{kl})\,\delta_{kl} \cr
  &= (5\,\delta_{kl})\,\delta_{kl} \cr
  &= 15 \cr
}$$
and the contraction $\big\{I:(\hat{n}\,\hat{n}\,\hat{n}\,\hat{n}):I\big\}$ yields 
$$\eqalign{
(\hat{n}\cdot\hat{n})\,(\hat{n}\cdot\hat{n}) &= (1)\,(1) = 1
}$$
So
$$\eqalign{
  15\,\beta &= \oint dA = 4\pi R^2 \cr
  \beta &= \frac{4\pi}{15} R^2 \cr
}$$
Another approach is to use Gauss' Theorem
$$\eqalign{
 \oint \hat{n}\,\hat{n}\,dA
  &= \frac{1}{R}\oint\vec{r}\,d\vec{a} \cr
  &= \frac{1}{R}\int\nabla\vec{r}\,dv \cr
  &= \frac{1}{R}\int\,I\,dv \cr
  &= \frac{4\pi}{3} R^2 I \cr
}$$
which matches our previous result.
On the other hand, the isotropic approach tells us immediately that 
$$\eqalign{
 \oint \hat{n}\,dA = 0 \cr
 \oint \hat{n}\,\hat{n}\,\hat{n}\,dA = 0 \cr
}$$
since in the first case, there is no isotropic vector. 
And in the second case,  there is an isotropic 3rd order tensor (i.e. the Levi-Civita permutation tensor), but it's not symmetric in all indices.
