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 1146: 18


$\dfrac{3 \pi}{2}$

Work Step by Step

$div F=\dfrac{\partial (z \tan^{-1} y^2)}{\partial x}+\dfrac{\partial (z^3(\ln (x^2+1))}{\partial y}+\dfrac{\partial (z)}{\partial z}=1$ $\iiint_Ediv \overrightarrow{F}dV=\int_{0}^{2 \pi}\int_0^{1} \int_{1}^{2-r^2} r dz dr d\theta$ or, $=2 \pi \times \int_{0}^{1} [zr] dr$ or, $=2 \pi \times \int_0^{1} r-r^3 dr$ or, $=\dfrac{\pi}{2}$ The flux through the disk is given by $=-\iint_{D} z dA=-\pi(1)^2=-\pi$ Hence the flux through the paraboloid is equal to the total Flux - Flux Through Disk Thus, we have $Flux=\dfrac{\pi}{2}-(-\pi)=\dfrac{3 \pi}{2}$
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