Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.7 - Surface Integrals - 16.7 Exercise - Page 1133: 7


$\dfrac{4 \sqrt 2-2}{3}$

Work Step by Step

Here we have $\iint_S y dS =\int_{0}^{\pi} \int_{0}^{1} (u \sin v) \sqrt {1+u^2} dA$ or, $=\int_{0}^{\pi} \sin v dv \int_{0}^{1} u [ \sqrt {1+u^2}] du$ or, $=\int_{0}^{1} 2u \sqrt {1+u^2} du$ $a=1+u^2$ and $2u du=da$ Now, $\iint_S y dS =\int_1^2 \sqrt {a} da$ or, $=\dfrac{2}{3}a^{3/2}$ Thus, we have $\iint_S y dS =\dfrac{4 \sqrt 2-2}{3}$
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