Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.7 Surface Integrals - 16.7 Exercises - Page 1173: 7


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

Work Step by Step

When the surface is orientable then the flux through a surface is defined as: $\iint_S F \cdot dS=\iint_S F \cdot n dS$ where, $n$ is the unit vector. Since, we have $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$ Consider $\iint_S y dS =\int_{0}^{\pi} \sin v dv \times \int_{0}^{1} u [ \sqrt {1+u^2}] du=\int_{0}^{1} 2u \times \sqrt {1+u^2} du$ Suppose $a=1+u^2; da=2u du$ $\iint_S y dS=\int_1^2 \sqrt {a} da=(\dfrac{2}{3}) (a^{3/2})$ Hence,$\iint_S y dS=\dfrac{4 \sqrt 2-2}{3}$
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