University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 14 - Section 14.3 - Area by Double Integration - Exercises - Page 772: 9

Answer

$$4$$

Work Step by Step

Our aim is to integrate the integral as follows: $ Area =\iint_D dA $ $\int^{2}_{0} \int^{3y}_{y} (1) \space dx \space dy = \int^{2}_{0}[x]^{3y}_y \space dy $ or, $=\int^{2}_{0}(2y)dy $ or, $=[y^2]^2_0$ So, $ Area=4$
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