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: 9


$171 \sqrt {14}$

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 $\iint_S f(x,y,z) dS \approx \Sigma_{i=1}^n f[\overline{x}, \overline{y}, \overline{z}] AS_i$ Since, $\iint_S x^2 yz dS =\iiint_{D}x^2 y (1+2x+3y) \times \sqrt {(2)^2+(3)^2+(1)^2} dA$ $\iint_S x^2 yz dS= \int_{0}^{3} \int_0^2 x^2y+2x^3y+3x^2y^2 dydx \times \sqrt {14}= \int_{0}^{3}[\dfrac{x^2 \times y^2}{2}+x^3 \times y^2+x^2 \times y^3]_0^2 dx \times \sqrt {14}$ or, $= \int_{0}^{3} 2x^2 +4x^3+8x^2 dx \times \sqrt {14}$ or, $=[\dfrac{10(3)^3}{3}+81] \times \sqrt {14}$ Hence, we have $\iint_S x^2 yz dS=171 \sqrt {14}$
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