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 1172: 4


$-80 \pi$

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. Since, we have $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$ Since, $\iint_S f(x,y,z) dS =g(\sqrt {x^2+y^2+z^2})$ This implies that $g(\sqrt 4) =g(2)=-5$ Now, we have $\iint_S f(x,y,z) dS =-5 \times \iint_S dS$ The integral $\iint_S dS$ shows the surface area of the sphere $=4 \pi r^2=4 \times \pi \times (4)=16 \pi$ Thus, we have $\iint_S f(x,y,z) dS =-5 \times 16 \pi=-80 \pi$
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.