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: 70


$2 \pi$

Work Step by Step

Consider $I= \int_{-1}^{1} \int_{-1/\sqrt {1-x^2}}^{1/\sqrt {1-x^2}} (2y+1) dy dx$ or, $= \int_{-1}^{1}[y^2+y] dx$ or, $=[2 \sin^{-1} x] _{-1}^{1}$ Thus, we have $I=2 \arcsin (1)-2 \arcsin (-1) =\pi+\pi=2 \pi$
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