Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.3 - Double Integrals in Polar Coordinates - 15.3 Exercise - Page 1014: 10

Answer

$$ \begin{aligned} \iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A =\frac{\pi}{2}\left(b^{2}-a^{2}\right) \end{aligned} $$

Work Step by Step

$$ \iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A $$ The region R can be described as $$ R=\left\{(x, y) | a \leqslant x^{2}+y^{2} \leqslant b \right\} $$ and in polar coordinates it is given by $$ R=\left\{(r, \theta) | a \leq r \leq b , \quad 0 \leq \theta \leq 2\pi \right\} $$ Therefore, $$ \begin{aligned} \iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A &=\int_{0}^{2 \pi} \int_{a}^{b} \frac{(r \sin \theta)^{2}}{r^{2}} r d r d \theta \\ &=\left(\int_{0}^{2 \pi} \sin ^{2} \theta d \theta\right)\left(\int_{a}^{b} r d r\right) \\ &=\int_{0}^{2 \pi} \frac{1}{2}(1-\cos 2 \theta) d \theta \int_{a}^{b} r d r \\ & =\frac{1}{2}\left[\theta-\frac{1}{2} \sin 2 \theta\right]_{0}^{2 \pi}\left[\frac{1}{2} r^{2}\right]_{a}^{b} \\ &=\frac{1}{2}(2 \pi-0-0)\left[\frac{1}{2}\left(b^{2}-a^{2}\right)\right] \\ &=\frac{\pi}{2}\left(b^{2}-a^{2}\right). \end{aligned} $$
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