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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 8



Work Step by Step

$\int^{1}_{-1} \int^{y^2-1}_{2y^2-2}dxdy$ =$\int^{1}_{-1}(y^2-1-2y^2+2)dy$ =$\int^{1}_{-1}(1-y^2)dy$ =[$y-\frac{y^3}{3}]^1_{-1}$ =$\frac{4}{3}$
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