University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.4 - Double Integrals in Polar Form - Exercises - Page 778: 12

Answer

$\pi a^2$

Work Step by Step

Conversion of polar coordinates and Cartesian coordinates are as follows: a)$r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$ b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ c) $x=r \cos \theta$ d) $y=r \sin \theta$ This information suggests that the integral over a circle of $4$ quadrants is as follows: $(2) \int_0^{a} \int_{{\frac{-\pi}{2}} }^{\frac{\pi}{2}} \int_0^{2} r dr d\theta=(2) \int_{{\frac{-\pi}{2}} }^{\frac{\pi}{2}} \dfrac{r^2}{2} d\theta$ Thus, $(2) \int_{{\frac{-\pi}{2}} }^{\frac{\pi}{2}} \dfrac{r^2}{2} d\theta=2(\dfrac{a^2}{2})\pi=\pi a^2$
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