Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.7 - Surface Integrals - 16.7 Exercise - Page 1192: 1

Answer

$ \approx -6.93$

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, $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f(\overline(x), \overline(y), \overline(z)) AS_i$ Here, $\iint_S F(x,y,z) dS =4[f(0,0,1) +f(0,1,0) +f(1,0,0)+f(-1,0,0) +f (0,-1,0) +f(0,0,-1)]$ $=4[-0.9899925-0.41614684+0.54030231+0.54030231-0.41614684-0.9899925]$ $ \approx -6.93$
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