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: 8

Answer

$\dfrac{384 \pi}{5}$

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 $div F=\dfrac{\partial (x^3+y^3)}{\partial x}+\dfrac{\partial (y^3+z^3)}{\partial y}+\dfrac{\partial (z^3+x^3)}{\partial z}=3x^2+3y^2+3z^2$ We get the spherical coordinates: $dV=dxdydz=\rho^2 \sin \phi d \rho d\theta d\phi$ $\iiint_E (3x^2+3y^2+3z^2)dV=\int_{0}^{\pi}\int_0^{2 \pi} \int_0^{2} (3 \rho^2) \rho^2 \sin \phi d\rho d\theta d\phi$ $=\int_{0}^{\pi}\int_0^{2 \pi} \int_0^{2} (3 \rho^4) \sin \phi d\rho d\theta d\phi$ $=3 [-\cos \phi]_{0}^{\pi} [\theta]_0^{2 \pi} [\dfrac{\rho^5}{5}]_0^2$ $=(3)\times (2) \times (2 \pi) \times \dfrac{32}{5}$ $=\dfrac{384 \pi}{5}$
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