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: 1

Answer

$\dfrac{9}{2}$

Work Step by Step

The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $ Here, we have $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV =\int_0^1\int_0^1\int_0^1 (3x+3) dz dy dx $ $=\int_0^1\int_0^1 [3xz+3z]_0^1 dy dx $ $=\int_0^1\int_0^1 3x(1)+3(1)-0 dy dx $ $=\int_0^1\int_0^1 (3x+3) dy dx $ $=\int_0^1 (3x+3)dx $ $=[(3/2)x^2+3x]_0^1$ $=\dfrac{9}{2}$
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