## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 15: Multiple Integrals - Section 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 13

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

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