Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.7 Exercises - Page 1145: 33

Answer

$\approx 4.5822$

Work Step by Step

The flux through a surface can be defined only when the surface is orientable. We know that $\iint_S F \cdot dS=\iint_S F \cdot n dS$ Here, $n$ denotes the unit vector. Since, $\iint_S (x^2+y^2+z^2) dS =\int_{0}^{1} \int_0^1 (x^2+y^2+z^2) \sqrt{1+(dx/dt)^2+ (dy/dt)^2} dA$ $=\int_{0}^{1} \int_0^1 (x^2+y^2+z^2) \sqrt{1+e^{2y}+x^2 e^{2y}} dx dy$ $=\int_{0}^{1} \int_0^1 (x^2+y^2+x^2e^{2y}) \sqrt{1+e^{2y}+x^2 e^{2y}} dx dy$ By using calculator tool, we have $ \approx 4.5822$
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