Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 60


$\dfrac{9 \pi-8}{3}$

Work Step by Step

Consider $A= \int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} (3-y) dy dx$ or, $= \int_{0}^{2} [3y-\dfrac{y^2}{2}]_{0}^{\sqrt{4-x^2}} dx$ or, $= \int_{0}^{2} [3 \sqrt {4-x^2} -\dfrac{4-x^2}{2}]dx$ or, $= [\dfrac{3\sqrt {4-x^2}x}{2}-2x +\dfrac{x^2}{3}+6 \sin^{-1} (\dfrac{x}{2})] _0^2 $ Thus, we have $I=6( \dfrac{\pi}{2})-4+\dfrac{8}{6}=\dfrac{9 \pi-8}{3}$
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