Answer
$\iint_S F\cdot dS=0$
Work Step by Step
Given: $G(x,y,z)=xi+yj+zk$
Then, we have $div G=1+1+1=3$
Consider $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$
Now, we have $\nabla f=f_xi+f_y j+f_z k=[\dfrac{-3x}{(x^2+y^2+z^2)^{5/2}}]i+[\dfrac{-3y}{(x^2+y^2+z^2)^{5/2}}]j+[\dfrac{-3z}{(x^2+y^2+z^2)^{5/2}}]k=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) $
Consider the rule such as: $div F=div (f G)=\nabla f \cdot G+f div G$
Then, we have $div (F)=div (f G)=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \cdot (xi+yj+zk) +\dfrac{3}{(x^2+y^2+z^2)^{3/2}}$
or, $-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(x^2+y^2+z^2)^2 +\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=-\dfrac{3}{(x^2+y^2+z^2)^{3/2}}+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}$
Divergence's Theorem states that $\iint_S F\cdot dS=\iiint_E div F dV$
$\iint_S F\cdot dS=\iiint_E div F dV=0$
Therefore, $\iint_S F\cdot dS=0$
