Answer
$0$
Work Step by Step
Divergence's Theorem states that $\iint_S F\cdot dS=\iiint_E div F dV$
Here, $S$ is a closed surface and $E$ is the region inside that surface.
Here, we have $G(x,y,z)=xi+yj+zk \implies div G=1+1+1=3$
Consider $f(x,y,z)=\dfrac{1}{(x^2+y^2+z^2)^{3/2}}$
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}}=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) $
Next, we have
$div F=div (f G)=\nabla f \cdot G+f div G$
or, $=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(xi+yj+zk) \cdot (xi+yj+zk) +\dfrac{3}{(x^2+y^2+z^2)^{3/2}}$
or, $=-\dfrac{3}{(x^2+y^2+z^2)^{5/2}}(x^2+y^2+z^2)+\dfrac{3}{(x^2+y^2+z^2)^{3/2}}$
Hence,
$\iint_S F\cdot dS=\iiint_E div F dV=0$
Therefore, $\iint_S F\cdot dS=0$
