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: 7


$\dfrac{9 \pi}{2}$

Work Step by Step

The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}+\dfrac{\partial c}{\partial z}$ This implies that $div F=\dfrac{\partial (3xy^2)}{\partial x}+\dfrac{\partial (xe^z)}{\partial y}+\dfrac{\partial (z^3)}{\partial z}=3y^2+0+3z^2=3(y^2+z^2)$ $\iiint_E 3(y^2+z^2)dV=\int_{-1}^{2}\int_0^{2 \pi} \int_0^{1} 3 (r^2 \cos^2 \theta+r^2 \sin^2 \theta) r dr d\theta dx$ $=\int_{-1}^{2} dx \times \int_0^{2 \pi} d\theta \times \int_0^{1} 3 r^3 dr$ $=[x]_{-1}^{2} \times [\theta]_0^{2 \pi} \times [3r^4/4]_0^{1} $ $=\dfrac{9 \pi}{2}$
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