Answer
4
Work Step by Step
$\int^{2}_{0} \int^{3y}_{y} 1 dx dy$
=$\int^{2}_{0}[x]^{3y}_ydy$
=$\int^{2}_{0}(2y)dy$
=$[y^2]^2_0$
=4
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: 9
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