Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.6 - Surface Integrals - Exercises 16.6 - Page 1001: 26


$\pi $

Work Step by Step

$F \cdot n =\dfrac{2xy^2 }{\sqrt {x^2+y^2}}+4xy +1 $ We set up and solve the surface integral as follows: $\iint_{S} F \cdot n \ dS=\iint_{R}\dfrac{2xy^2 }{\sqrt {x^2+y^2}}+4xy +1 \ dy \ dx$ Now, we will convert the integral into polar coordinates. $I=\int_0^{2 \pi} \int_{0}^{1} \dfrac{2 (r \cos \theta)(r \sin \theta)^2 }{\sqrt {r^2}}+4 (r \cos \theta)(r \sin \theta) +1 $ or, $=\int_0^{2 \pi} \int_{0}^{1} (2r^3 \sin^2 \theta \cos \theta+4r^3 \sin \theta \cos \theta +r) dr d \theta$ or, $= \int_0^{2 \pi} [\dfrac{1}{2} \sin^2 \theta \cos \theta +\sin \theta \cos \theta+\dfrac{1}{2}] \ d \theta$ or, $=\pi $
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