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: 10



Work Step by Step

Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $ where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}=\dfrac{\partial (z)}{\partial x}+\dfrac{\partial (y)}{\partial y}+\dfrac{\partial (zx)}{\partial z}=1+x$ $\iiint_E (1+x)dV=\int_{0}^{a}\int_0^{b(1-(x/a))} \int_0^{c(1-(x/a)-(y/b))} (1+x) dzdydx$ $=\int_{0}^{a}\int_0^{b(1-(x/a))} (1+x) \times (z)_0^{c(1-(x/a)-(y/b))}dydx$ Suppose $A=1-\dfrac{x}{a} $ and $dA=\dfrac{-dx}{a}$ and $ dx=-adA$ $\iiint_E (1+x)dV=\int_{0}^{1} (ac) \times (1+a-aA)[\dfrac{bA^2}{2}]dA=\dfrac{1}{2}(abc) \times \int_{0}^{1} (A^2+aA^2-aA^3) dA$ $=\dfrac{1}{2}(abc) [\dfrac{A^3}{3}+\dfrac{aA^3}{3}-\dfrac{aA^4}{4}]_0^1 du$ $=\dfrac{1}{2} \times (abc) \times \dfrac{1}{3}+\dfrac{a}{3}-\dfrac{a}{4}$ $=\dfrac{(a+4)abc}{24}$
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