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 1133: 9

Answer

$171 \sqrt {14}$

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 orientable. Here,we have $\iint_S x^2 yz dS =\iiint_{D}x^2 y (1+2x+3y) \sqrt {2^2+3^2+1} dA=\iiint_{D}x^2 y (1+2x+3y) \sqrt {4+9+1} dA$ or, $= \sqrt {14} \times \int_{0}^{3} \int_0^2 x^2y+2x^3y+3x^2y^2 dydx$ or, $= \sqrt {14} \times \int_{0}^{3}[\dfrac{x^2y^2}{2}+x^3y^2+x^2y^3]_0^2 dx$ or, $= \sqrt {14} \times \int_{0}^{3} 2x^2 +4x^3+8x^2 dx$ Thus, we have $\iint_S x^2 yz dS=(\sqrt {14}) [\dfrac{10(3)^3}{3}+81)=171 \sqrt {14}$
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