Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Exercises - Page 675: 31

Answer

$A=\pi ab$

Work Step by Step

The ellipse is symmetric about the axes, so we find the area in the first quadrant and multiply it by 4. $A=4\displaystyle \int_{0}^{a}ydx=\quad \left[\begin{array}{ll} \frac{dx}{d\theta}=-a\sin\theta & \\ x=0\Rightarrow\theta=\frac{\pi}{2}, & x=a\Rightarrow\theta=0 \end{array}\right]$ $=4\displaystyle \int_{\pi/2}^{0}b\sin\theta(-a\sin\theta)d\theta$ $=4ab\displaystyle \int_{\pi/2}^{0}-\sin^{2}\theta d\theta$ $=4ab\displaystyle \int_{0}^{\pi/2}\sin^{2}\theta d\theta\qquad...\left[\begin{array}{l} \cos 2\theta=1-2\sin^{2}\theta\\ \sin^{2}\theta=(1-\cos 2\theta)/2 \end{array}\right]$ $=4ab\displaystyle \int_{0}^{\pi/2}\frac{1}{2}(1-\cos 2\theta)d\theta$ $=2a[\displaystyle \int_{0}^{\pi/2}d\theta-\int_{0}^{\pi/2}\cos 2\theta d\theta]$ $=2ab[\displaystyle \theta-\frac{1}{2}\sin 2\theta]_{0}^{\pi/2}$ $=2ab[(\displaystyle \frac{\pi}{2}-0)-(0-\frac{1}{2}(0)]$ $=2ab(\displaystyle \frac{\pi}{2})$ $=\pi ab$
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