## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 882: 20

#### Answer

$\dfrac{\pi}{4}$

#### Work Step by Step

We rewrite as: $I= \int_0^{\pi} \int_0^{\sin x} y dy dx$ or, $= \int_0^{\pi} [y^2/2]_0^{\sin x} dx$ Use formula: $\sin ^2 x= \dfrac{1- \cos 2 x}{2}$ Now, $I=(1/4) \int_0^{\pi} 1-\cos 2x dx$ or, $=(1/4) [\pi -\dfrac{\sin 2 \pi}{2} ]$ So, $I=\dfrac{\pi}{4}$

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