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 1173: 33

Answer

$\approx 4.5822$

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 $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ Since, $\iint_S (x^2+y^2+z^2) dS =\int_{0}^{1} \int_0^1 (x^2+y^2+z^2) \times \sqrt{1+(\dfrac{dx}{dt})^2+ (\dfrac{dy}{dt})^2} dA$ and $\iint_S (x^2+y^2+z^2) dS =\int_{0}^{1} \int_0^1 (x^2+y^2+z^2) \sqrt{1+e^{2y}+x^2 \times e^{2y}} dx dy=\int_{0}^{1} \int_0^1 (x^2+y^2+x^2 \times e^{2y}) \sqrt{1+e^{2y}+x^2 \times e^{2y}} dx dy$ Need to use calculator tool. $\iint_S (x^2+y^2+z^2) dS \approx 4.5822$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.