Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.4 Integration in Polar, Cylindrical, and Spherical Coordinates - Exercises - Page 880: 16

Answer

$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = 0$

Work Step by Step

We have $f\left( {x,y} \right) = x$ and the region $2 \le {x^2} + {y^2} \le 4$. In polar coordinates we have $f\left( {r\cos \theta ,r\sin \theta } \right) = r\cos \theta $ The description implies that the region is an annulus located between the disk of radius $\sqrt 2 $ and the disk of radius $2$. Thus, the region description in polar coordinates: ${\cal D} = \left\{ {\left( {r,\theta } \right)|\sqrt 2 \le r \le 2,0 \le \theta \le 2\pi } \right\}$ Using the Change of Variables Formula for ${\cal D}$, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {r\cos \theta ,r\sin \theta } \right)r{\rm{d}}r{\rm{d}}\theta $ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt 2 }^2 \left( {r\cos \theta } \right)r{\rm{d}}r{\rm{d}}\theta $ $ = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = \sqrt 2 }^2 {r^2}{\rm{d}}r} \right)$ $ = \left( {\sin \theta |_0^{2\pi }} \right)\left( {\frac{1}{3}{r^3}|_{\sqrt 2 }^2} \right) = 0$ Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = \sqrt 2 }^2 \left( {r\cos \theta } \right)r{\rm{d}}r{\rm{d}}\theta = 0$.
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