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: 11


$\dfrac{\sqrt {21}}{3}$

Work Step by Step

Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$ where, $n$ denotes the unit vector. and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ The flux through a surface can be defined only when the surface is orientable. Here, we have $\iint_S x dS =\iiint_{D}x \sqrt {(-4)^2+(2)^2+1^2} dA$ or, $=\sqrt{21} \times \int_{0}^{1} \int_{2x-2}^0 x dy dx$ or, $=\sqrt{21} \times \int_{0}^{1} [xy]_{2x-2}^0 \times x dx$ or, $=- \sqrt{21} \times \int_{0}^{1} (2x^2-2x) dx$ or, $=- \sqrt{21} \times [(]\dfrac{2x^3}{3})-x^2]_{0}^{1}$ or, $\iint_S x dS=-\sqrt {21} [(\dfrac{2(1)^3}{3}-0)-[(1)^2-0]]_0^1=\dfrac{\sqrt {21}}{3}$
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