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: 10



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 x z dS =\iiint_{D}x(4-2x-2y) \sqrt {(-2)^2+(-2)^2+1} dA= \int_{0}^{2} \int_0^{2- x} 4x-2x^2 -2xy dy dx \times (3) $ $\iint_S x z dS= \int_{0}^{2} (4x-2x^2) \times [(2-x) -x (2-x)^2] \times (3) dx= \int_{0}^{2} x(2-x)^2 dx \times 3 $ Hence, we have $\iint_S x z dS=\int_0^2 2x^2-x^3 dx \times 3 =[\dfrac{2x^3}{3}-\dfrac{x^4}{4}]_0^2=\dfrac{2(2)^3}{3}-\dfrac{(2)^4}{4}=4$
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