Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.2 - Double Integrals over General Regions - 15.2 Exercise - Page 1062: 79

Answer

$\pi a^2 b$

Work Step by Step

Consider $I=\iint_{D} (ax^3+by^3+\sqrt {a^2-x^2}) d A$ or, $=\int_{-a}^a \int_{-b}^b (ax^3+by^3+\sqrt {a^2-x^2}) dy \ dx $ or, $=\int_{-a}^a (ayx^3+\dfrac{by^4}{4}+y\sqrt {a^2-x^2})_{-b}^b \ dx $ or, $= 2b \int_{-a}^a ax^3 \ dx + 2b \int_{-a}^a \sqrt {a^2-x^2} \ dx $ or, $=2b \int_{-a}^a \sqrt {a^2-x^2} \ dx $ or, $=2b [\dfrac{u\sqrt {a^2-u^2}}{2} +\dfrac{a^2}{2} \sin^{-1} \dfrac{u}{a}]_{-a}^{a}$ or, $=4b[\dfrac{a\sqrt {a^2-a^2}}{2} +\dfrac{a^2}{2} \sin^{-1} \dfrac{a}{a}]$ or, $=2a^2b \sin^{-1} (1) $ or, $=2a^2b \times \dfrac{\pi}{2}$ or, $=\pi a^2 b$
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