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: 21



Work Step by Step

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