For $x \, \epsilon \,R^n$, prove $\int_{|x| \leq 1} x_i x_j dx = 0$

Question

Let $x \, \epsilon \, R^n$, with $x_i$ corresponding coordinates. Then $|x| \leq 1$ is the unit ball in $R^n$. How can I easily prove that $\int_{|x| \leq 1} x_i x_j dx = 0$ if $i \ne j$? It is kind of intuitive, but I don't seem to see it through.

Answer 1

$A = \int_{|x| \leq 1} x_i x_j dx$

$= - \int_{|x| \leq 1} (-x_i) x_j dx$

$= - \int_{|x| \leq 1} x_i x_j dx$ (because $|x| \leq 1$ is symmetric around $x_i = 0$)

$ = - A$

Hence $A = 0$.

Answer 2

A little overpowered: $T_{ij}=\int_{|x|<1} x_{i}x_{j} dV$ is clearly isotropic (the sphere is invariant under rotations; just verify the transformation rule). Therefore $T_{ij}=\alpha \delta_{ij}$ for some scalar $\alpha$.