Integral of dyadic double dot product on a unit sphere

Question

I am trying to reconstruct an analysis from fluid dynamics textbook and I got all terms with the exception of one:

$$ \oint_S\! (\mathbf{e} : (\mathbf{n} \, \mathbf{n}))\, \mathbf{n}\, \mathbf{n}\, dA, $$ where $\mathbf{e}$ is a 2nd order symmetric tensor with $e_{ii} = 0$ and $\mathbf{n}$ are normals. The integration is over a unit sphere.

I believe the result should be 0. The non-matching indices in the contraction will form odd functions so vanish under integral whereas matching indices will vanish due do $e_{ii}=0$ condition. But I'd like a confirmation or some neater proof of this. I don't really have geometric intuition about double dot products of dyadics so I am afraid my operational transformations may be misguided.

Answer 1

Assuming that $e$ is constant, move it outside of the integral $$J = e:\oint n\,n\,n\,n\,dA$$ Then use the answer to this question to write it as $$\eqalign{ J &= \frac{4\pi R^2}{15}\,(e:{\mathbb Y}) \cr &= \frac{4\pi R^2}{15}\,(e+e^T+e\operatorname{tr}(e)) \cr &= \frac{4\pi R^2}{15}\,(2e) \cr &= \frac{8\pi R^2}{15}\,e \cr }$$