Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Review - Exercises - Page 1150: 36



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, $\iint_S F\cdot dS=0$
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