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: 22

Answer

The divergence is positive for the points above the region $x+y=0$ and negative for the points below the region $x+y=0$

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. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}=2x+2y=2(x+y)$ Hence, the divergence is positive for the points above the region $x+y=0$ and negative for the points below the region $x+y=0$.
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