Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.9 The Divergence Theorem - 16.9 Exercises - Page 1185: 13


$2 \pi$

Work Step by Step

Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=4 \sqrt {x^2+y^2+z^2}$ Consider the integral $I=(4) \int_{0}^{2 \pi}\int_0^{\pi/2} \int_{0}^{1} \sqrt {x^2+y^2+z^2} dV$ $=(4)\int_{0}^{2 \pi}\int_0^{\pi/2} \int_{0}^{1} \sqrt{\rho^2} \rho^2 \sin \phi \times d \rho d\phi d \theta$ $=(4) [\int_{0}^{2 \pi} d\theta] [\int_0^{\pi/2} \sin \phi d\phi] [\int_{0}^{1}[\rho^3 d \rho] $ $=(8 \pi) \times (-\cos \phi)_0^{\pi/2} \times [\dfrac{\rho^4}{4}]_0^1$ Hence, $I=2 \pi$
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