Calculus: Early Transcendentals 8th Edition

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

Chapter 16 - Section 16.9 - The Divergence Theorem - 16.9 Exercise - Page 1146: 31

Answer

$\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$

Work Step by Step

Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ Here, $S$ shows a closed surface and $E$ is the region inside that surface. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$ Here, we have $\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$ Re-arrange as: $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$ This implies that $\iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$ and $\iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$ Thus, we have $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$ Hence, our result has been proved.
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