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: 24



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. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=\dfrac{\partial (2)}{\partial x}+\dfrac{\partial (2)}{\partial y}+\dfrac{\partial z}{\partial z}=0+0+1=1$ The integral $\iiint_E div F dV$ is defined as the volume of the region $E$ that lies inside a sphere having radius $1$. The volume of the region E is: $\iiint_E dV=\dfrac{4\pi(1)^3}{3}=\dfrac{4\pi}{3}$ Hence, we have: $\iint_S F \cdot n dS=\iiint_Ediv \overrightarrow{F}dV =\dfrac{4\pi}{3}$
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