Surface Integral 2

The general vector surface area element $d\mathbf{S}$ can be written in components as $(dy\,dz,dx\,dz,dx\,dy)$ , and so the integral is $I \; =\; \iint_\Omega \frac{x\,dydz + y\,dzdx + z\,dxdy}{(x^2 + y^2 + z^2)^{\frac32}} \; = \; \iint_\Omega \frac{\mathbf{r}}{r^3}\,d\mathbf{S} \; = \; -\iint_\Omega \nabla \tfrac{1}{r}\,d\mathbf{S}$ Since $r^{-1}$ is Harmonic on $\mathbb{R}^3 \backslash\{\mathbf{0}\}$ , Green's Theorem tells us that $I \; = \; -\iint_{C_\epsilon} \nabla \tfrac{1}{r}\,d\mathbf{S}$ where $C_\epsilon$ is the sphere with radius $\epsilon$ centred at the origin, and hence $I \; = \; \iint_{C_\epsilon} \frac{dS}{r^2} \; = \; \iint_{C_\epsilon} \frac{\epsilon^2\,d\Omega}{\epsilon^2} \; = \; \boxed{4\pi}$ where the final integral $d\Omega$ is with respect to solid angle.