A second-rank tensor $T(\mathbf{y})$ is defined by

$$T_{i j}(\mathbf{y})=\int_{S}\left(y_{i}-x_{i}\right)\left(y_{j}-x_{j}\right)|\mathbf{y}-\mathbf{x}|^{2 n-2} \mathrm{~d} A(\mathbf{x})$$

where $\mathbf{y}$ is a fixed vector with $|\mathbf{y}|=a, n>-1$, and the integration is over all points $\mathbf{x}$ lying on the surface $S$ of the sphere of radius $a$, centred on the origin. Explain briefly why $T$ might be expected to have the form

$$T_{i j}=\alpha \delta_{i j}+\beta y_{i} y_{j}$$

where $\alpha$ and $\beta$ are scalar constants.

Show that $\mathbf{y} \cdot(\mathbf{y}-\mathbf{x})=a^{2}(1-\cos \theta)$, where $\theta$ is the angle between $\mathbf{y}$ and $\mathbf{x}$, and find a similar expression for $|\mathbf{y}-\mathbf{x}|^{2}$. Using suitably chosen spherical polar coordinates, show that

$$y_{i} T_{i j} y_{j}=\frac{\pi a^{2}(2 a)^{2 n+2}}{n+2}$$

Hence, by evaluating another scalar integral, determine $\alpha$ and $\beta$, and find the value of $n$ for which $T$ is isotropic.