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: 27


$\dfrac{-73 \pi}{6} $

Work Step by Step

$F \cdot n =\dfrac{-x^2 }{\sqrt {x^2+y^2}}-\dfrac{y^2 }{\sqrt {x^2+y^2}} -x^2-y^2$ We set up and solve the surface integral as follows: $\iint_{S} F \cdot n \ dS=\iint_{R} [\dfrac{-x^2 }{\sqrt {x^2+y^2}}-\dfrac{y^2 }{\sqrt {x^2+y^2}} -x^2-y^2] \ dy \ dx$ Now, we will convert the integral into polar coordinates. $I=-\int_0^{2 \pi} \int_{1}^{2} [\dfrac{r^2 \cos^2 \theta}{\sqrt {r^2}}+\dfrac{r^2 \sin^2 \theta)}{(\sqrt {r^2}}+r^2 \cos^2 \theta +r^2 \sin^2 \theta] r dr d \theta$ or, $=-\int_0^{2 \pi} \int_{1}^{2} (r^2+r^3) r dr d \theta$ or, $= -\int_0^{2 \pi} [\dfrac{1}{3} r^3 +\dfrac{1}{4} r^4]_1^2 \ d \theta$ or, $=\dfrac{-73 \pi}{6} $
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