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: 63

Answer

$20$

Work Step by Step

Consider $I= \iint_{R} f(x,y) dA$ or, $= \int_{0}^{2} \int_{0}^{2-x} (12-3y^2) dy dx$ or, $= \int_{0}^{2}[12y-y^3]_0^{2-x} dx$ or, $=\int_{0}^{2} (24-12x-(2-x)^3) dx$ Thus, we have $I=[24 x-6x^2+ \dfrac{(2-x)^4}{4}]_0^2=20$
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