Answer
$\frac{4}{3}$
Work Step by Step
$\int^{1}_{-1} \int^{y^2-1}_{2y^2-2}dxdy$
=$\int^{1}_{-1}(y^2-1-2y^2+2)dy$
=$\int^{1}_{-1}(1-y^2)dy$
=[$y-\frac{y^3}{3}]^1_{-1}$
=$\frac{4}{3}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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