Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.9 - The Divergence Theorem - 16.9 Exercise - Page 1206: 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 $ $=[(3/2)x^2+3x]_0^1$ $=\dfrac{9}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.