Calculus 8th Edition

$\iint_S F\cdot dS=0$
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, $\iint_S F\cdot dS=0$