Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.9 - The Divergence Theorem - 16.9 Exercise - Page 1207: 21

Answer

$P_1$ is negative and $P_2$ is positive.

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. $div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$ When the net flow of water is towards a point, then the divergence at that point is negative, but when the net flow of water is outwards, then the divergence at that point is positive. If there is no net flow of water, then the divergence at that point will be zero. It has been seen that at the point $P_1$ the net flow of water is inwards and hence the divergence at that point $P_1$ is negative. At the point $P_2$, the net flow of water is outwards and thus the divergence at $P_2$ is positive. Hence, $P_1$ is negative and point $P_2$ is positive.
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