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: 21

Answer

$4$

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 and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ Consider $\iint_S F \cdot n dS=\iint_D 2 (u^2-v^2) dA=2 \times \int_{0}^{1} \int_0^{2} 2 (u^2-v^2) du dv$ or, $\iint_S F \cdot n dS=(2) \int_{0}^{1} [\dfrac{u^3}{3}-uv^2]_0^2 dv= \int_{0}^{1} \dfrac{2^3}{3}-(2)v^2 dv \times $ Hence, we have $\iint_S F \cdot n dS=[\dfrac{8}{3}-\dfrac{2}{3}] \times (2)=4$
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