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: 8


$\pi \sqrt 2$

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. Consider $\iint_D f(r(u,v)) y dS =\iiint_{S}[4u^2v^2 +(u^2+v^2)^2] \times ( 4 \sqrt 2 (u^2+v^2) dA$ or, $\iint_D f(r(u,v)) y dS =\int_{0}^{2 \pi} \int_0^1 (4r^4 \cos^2 \theta \sin^2 \theta +(r^2 \cos^2 \theta -r^2 \sin^2)^2] r^3 dr d \theta \times (4 \sqrt 2) $ or, $\iint_D f(r(u,v)) y dS = \int_{0}^{2 \pi} \int_0^1 [r^4 \sin^2 (2 \theta) +(r^4 \cos^2 (2 \theta) r^3 dr d\theta \times (4 \sqrt 2)$ or, $\iint_D f(r(u,v)) y dS =\int_{0}^{2 \pi} \int_0^1 r^7 dr d\theta \times (4 \sqrt 2)$ Hence, we have $\iint_D f(r(u,v)) y dS =(\dfrac{\sqrt 2}{2}) \int_0^{2 \pi} d \theta=\pi \sqrt 2$
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