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: 34



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. Since, $\iint_S xyz dS =\iint_{D} xyz \sqrt{1+(dz/dx)^2+ (dz/dy)^2} dA$ $=\iint_{D} xyz \sqrt{1+(2xy^2)^2+(2x^2y)^2} dx dy$ $=\int_0^2 \int_0^1 xyz \sqrt{1+4x^2y^4+ 4x^4y^2} dx dy$ $=\int_0^2 \int_0^1 x^3y^3 \sqrt{1+4x^2y^4+ 4x^4y^2} dx dy$ Now use a calculating tool. $ \approx 4.92429$
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