## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 15: Multiple Integrals - Section 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 19

#### Answer

a)0 b)$\frac{4}{\pi}$

#### Work Step by Step

a) average =$\frac{1}{\pi} \int^{\pi}_{0} \int^{\pi}_{0} sin(x+y)dy dx$ =$\frac{1}{\pi^2} \int^{\pi}_{0} [-cos(x+y)]^{\pi}_0dx$ =$\frac{1}{\pi^2}\int^{\pi}_{0} [-cos(x+\pi)+cosx]dx$ =$\frac{1}{\pi^2}[-sin(x+\pi)+sin x]^\pi_0$ =$\frac{1}{\pi}[(-sin 2\pi+sin \pi)-(-sin \pi +sin0)]$ =0 --- b) average=$\frac{1}{\frac{\pi^2}{2}}\int^{\pi}_{0} \int^{\pi/2}_{0} sin(x+y)dydx$ =$\frac{2}{\pi^2} \int^{\pi}_{0} [-cos(x+y)]^{\pi/2}_0dx$ =$\frac{2}{\pi^2}\int^{\pi}_{0} [-cos(x+\frac{\pi}{2})+cosx]dx$ =$\frac{2}{\pi^2}[-sin(x+\frac{\pi}{2})+sin x]^\pi_0$ =$\frac{2}{\pi}[(-sin \frac{3\pi}{2}+sin \pi)-(-sin \frac{\pi}{2}+sin 0)]$ =$\frac{4}{\pi^2}$

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