Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.7 - Surface Integrals - 16.7 Exercise - Page 1132: 3

Answer

$900 \pi$

Work Step by Step

Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$ where, $n$ denotes the unit vector. and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ The flux through a surface can be defined only when the surface is orient-able. Since, $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f(\overline(x), \overline(y), \overline(z)) AS_i$ The four parts of the surface consists of the same surface area and the area of each part is given by: $\dfrac{4 \pi r^2}{8}=\dfrac{50 \pi}{2}= 25\pi$ Thus, we have $\iint_S F(x,y,z) dS =25 \pi[7+8+9+12]=900 \pi$
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