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: 5

Answer

$\dfrac{9}{2}$

Work Step by Step

$div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ This implies that $div F=\dfrac{\partial (xye^z)}{\partial x}+\dfrac{\partial (xy^2z^3)}{\partial y}+\dfrac{\partial (-ye^z)}{\partial z}=ye^z+2xyz^3-ye^z=2xyz^3$ The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ $=\int_{0}^3\int_0^2 \int_0^1 (2xyz^3)dzdydx$ $=\int_{0}^3\int_0^2[\dfrac{xy}{2}]dydx$ $=\int_{0}^3 x dx$ $=[x^2/2]_0^3$ $=\dfrac{9}{2}$
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