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: 34


$\approx 4.92429$

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