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 882: 30



Work Step by Step

Consider $\iint_{R} f(s,t) dA= \int_{0}^{1} \int_{0}^{\sqrt{1-s^2}} 8t dt ds= \int_{0}^{1} [4t^2]_{0}^{\sqrt{1-s^2}} ds$ or, $= \int_{0}^{1} 4(1-s^2) ds$ or, $= [4s -\dfrac{4s^3}{3}]_0^1$ Thus, we have $I=4(1-0) -\dfrac{4}{3}=\dfrac{8}{3}$
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