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: 5


$11\sqrt {14}$

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 $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$ Consider $\iint_S f(x+y+z) dS =\int_{0}^2 \int_{0}^1 (4u+1+v) \times (\sqrt {14}) (dv du)=(\sqrt {14}) \int_{0}^2 [(4uv+v+\dfrac{v^2}{2}) du$ Now, $\iint_S f(x+y+z) dS=(\sqrt {14}) \int_{0}^2 [(4u+1+\dfrac{1}{2}) du=(\sqrt {14}) [2u^2+\dfrac{3u}{2}]_0^2$ Hence, we have $\iint_S f(x+y+z) dS=11\sqrt {14}$
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