Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 15

Answer

$2-\sqrt{3}$

Work Step by Step

$\int^{\sqrt{3}}_1 \int^x_1dydx $ =$\int^{\pi/4}_{\pi/6} \int^{\sqrt{3} \sec\theta}_{csc\theta}rdrd\theta $ =$\int^{\pi/4}_{\pi/6}(\frac{3}{2}sec^2\theta-\frac{1}{2}csc^2\theta) d\theta $ =$[\frac{3}{2}tan\theta+\frac{1}{2}cot\theta]_{\pi/6}^{\pi/4}$ =2-$\sqrt{3}$
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