Answer
$\iint_S F\cdot dS=0$
Work Step by Step
Apply Divergence's Theorem as: $\iint_S F\cdot dS=\iiint_E div F dV$
Here, $S$ shows a closed surface and $E$ is the region inside that surface $S$.
Given: $G(x,y,z)=xi+yj+zk$
and $ div G=1+1+1=3$
Let us consider that $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$
Now, $\nabla f=f_xi+f_y j+f_z k$
or, $\nabla f=\dfrac{-3x}{(x^2+y^2+z^2)^{5/2}}i+\dfrac{-3y}{(x^2+y^2+z^2)^{5/2}} j+\dfrac{-3z}{(x^2+y^2+z^2)^{5/2}}$
or, $\nabla f=-\dfrac{3(xi+yj+zk)}{(x^2+y^2+z^2)^{5/2}}$
Now,we have $div F=div (f G)=\nabla f \cdot G+f (div G)$
and
$-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \times (xi+yj+zk) +\dfrac{3}{(x^2+y^2+z^2)^{3/2}}=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(x^2+y^2+z^2)+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}$
This implies that $\iint_S F\cdot dS=\iiint_E div F dV=0$
Hence, $\iint_S F\cdot dS=0$
