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: 11



Work Step by Step

Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial (2x^3+y^3)}{\partial x}+\dfrac{\partial (y^3+z^3)}{\partial y}+\dfrac{\partial (3y^2z)}{\partial z}=6(x^2+y^2)$ $\iiint_E 6(x^2+y^2) dV=\int_{0}^{2 \pi}\int_0^{1} \int_0^{1-r^2} 6r^2 \times (r dz dr d\theta)=\int_{0}^{2 \pi}\int_0^{1} (6r^3 z)_0^{1-r^2} dr d\theta$ $=\int_{0}^{2 \pi}[\dfrac{3r^4}{2}-r^6]_{0}^{1} d\theta$ $=\int_{0}^{2 \pi} [\dfrac{3(1)^4}{2}-(1)^6] d\theta$ $=\int_{0}^{2 \pi}(\dfrac{3}{2}-1) d\theta$ Hence, we have $\iiint_E 6(x^2+y^2) dV=[\dfrac{\theta}{2}]_0^{2 \pi}=(\dfrac{ 2\pi}{2}-0)=\pi$
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