Answer
12
Work Step by Step
$\int^{6}_{0} \int^{2y}_{y^2/3}dxdy $
=$\int^{6}_{0} (2y-\frac{y^2}{3})dy$
=$[y^2-\frac{y^3}{9}^6_0]$
=$36-\frac{216}{9}$
=12
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: 13
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