Axisymmetric symmetric rank-two tensor ($T=\alpha I+\beta y{y}^T$)

A symmetric tensor invariant under rotations about a fixed nonzero vector $y$ has the form $\alpha I+\beta y{y}^T$. Choose the third coordinate axis parallel to $y$: rotation by a half-turn makes the mixed axial-transverse entries vanish, and all transverse rotations make the transverse block a scalar multiple of the identity. Its two transverse eigenvalues are $\alpha$ and its axial eigenvalue is $\alpha +\beta |y{|}^2$. The trace and the quadratic contraction $y^{T}Ty$ determine these two coefficients. Full isotropy holds exactly when $\beta =0$.

For $|y|=a$ and $n>-1$, this surface integral is finite: near $x=y$ its integrand has size $O(|x-y{|}^{2n})$ and the two-dimensional area element supplies one further power of distance. Rotational covariance and symmetry make it an axisymmetric symmetric rank-two tensor. Put $K=\pi (2a{)}^{2n+2}$. Polar integration about the $y$ axis gives $\mathrm{tr}T=K/(n+1)$ and $y^{T}Ty=a^{2}K/(n+2)$. Solving $3\alpha +\beta a^2=\mathrm{tr}T$ and $\alpha +\beta a^2=y^{T}Ty/a^2$ gives $\alpha =K/[2(n+1)(n+2)]$ and $\beta =\alpha (2n+1)/a^2$. Thus the weighted chord tensor becomes an isotropic tensor exactly at $n=-1/2$.