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 1145: 9



Work Step by Step

The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ Here, $S$ is a closed surface and $E$ is the region inside that surface. $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ This implies that $div F=\dfrac{\partial (xe^y)}{\partial x}+\dfrac{\partial (z-e^y)}{\partial y}+\dfrac{\partial (-xy)}{\partial z}$ $div F=e^y-e^y-0=0$ Now, we have $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV = \iiint_E (0) dV=0$ Thus, we have $\iiint_S F \cdot dS = 0$
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