Evaluate the integral $$\iint_S (z^2 + y^2 + x^2) \, dS ,$$ where $S$ is the surface of the cube $\{-a < x < a, -a < y< a, -a< z< a\}$.
I've attempted to partition the surface into its six faces and expressed the element of surface area dS for each face, but I'm uncertain about how to proceed further.
Any insights on integrating over each face and summing the results to obtain the final solution would be greatly appreciated.
Thank you in advance for any assistance provided.
Answer from book: $40a^4$
By symmetry of the integrand, $$\iint_S(x^2+y^2+z^2)\,\mathrm dS=24\int_0^a\int_0^a(x^2+y^2+a^2)\, \mathrm dx\,\mathrm dy=24\int_0^a\left(\frac{4}{3}a^3+ay^2\right)\mathrm dy=40a^2$$