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 1174: 47


$1248 \pi$

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 $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ Here, we have $r_s \times r_v=\sqrt 6 (\cos u j+\sin u k)$ and $dS=\pm (r_u \times r_v) dA=\sqrt 6 ( \cos u j+\sin u k) dA$ Here, $F(r(u,v))=26 \sqrt 6 (\cos u j+\sin u k)$ Here, $\iint_S F \cdot dS= \int_{0}^4 \int_{0}^{2 \pi} 26 \sqrt 6 \times [\cos (u) j+\sin (u) k] \times [\sqrt 6 ( \cos u j+\sin u k) dA] $ Hence, $\iint_S F \cdot dS=\int_{0}^4 \int_{0}^{2 \pi} 26 \times 6 \times (\cos^2 u+\sin^2 u) dA=(26) \cdot (6) \cdot (4) \cdot (2 \pi)=1248 \pi$
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