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: 61

Answer

$16$

Work Step by Step

Consider: $I= \iint_{R} f(x,y) dA$ or, $= \int_{0}^{2} \int_{0}^{3} (4-y^2) dx dy$ or, $= \int_{0}^{2}[4x-y^2 x]_0^{3} dy$ or, $=\int_{0}^{2} (12-3y^2) dy$ Thus, we have $I=[12y -y^3]_{0}^{2} =24-18=16$
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