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: 26


$V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$

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. $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ Thus, we have $div F=\dfrac{\partial x}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial z}{\partial z}=1+1+1=3$ $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E (3)dV $ $(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_E dV$ Hence, it has been verified that $V(E)=(\dfrac{1}{3}) \iint_S \overrightarrow{F}\cdot d\overrightarrow{S}$
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