How to Calculate this Surface Integral?

Question

Suppose that $i=j$ in $$\iint \limits_S\hat{n}_i\hat{n}_j \, dA = \iint \limits_S\hat{n}_i^2\,dA$$ where $S$ is the sphere centered at the origin with the radius $R$ and $\hat{n}_i$ and $\hat{n}_j$ are the $x$ and $y$-components of the vector $\hat{n}$ which is normal to the sphere. What is the value of this surface integral?

Answer 1

HINT: Note that by symmetry of the sphere the integrals of $\hat n_i^2$ are the same for $i=1,2,3$. But $\hat n_1^2+\hat n_2^2+\hat n_3^2 = \|\hat n\|^2 = 1$, so the sum of their integrals is the area of the sphere, $4\pi R^2$.

(If you want to do an explicit calculation, use spherical coordinates.)