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 894: 47


$\dfrac{8}{9} (3\pi-4)$

Work Step by Step

We calculate the double integral as follows: $ \iint{R} \sqrt {x^2+y^2} \ dA=\int^{0}_{\pi} \int_{2 \ sin\theta}^2{} \ r \ r dr \ d\theta \\=\int_{0}^{\pi} \int_{2 \ sin\theta}^2 r^2 dr \ d\theta \\=\int_{0}^{\pi} [\dfrac{r^3}{3}]_{2 \sin\theta}^2 d\theta \\=\int_{0}^{\pi} [\dfrac{8}{3}-\dfrac{8 \sin^3 \theta}{3}] \ d\theta \\=\dfrac{8}{3} \times \int_{0}^{\pi} (1- \sin^3 \theta) \ d \theta \\=\dfrac{8}{3} [\theta +\cos \theta -\dfrac{\cos^3 \theta }{3}]_0^{\pi} \\=\dfrac{8}{3} (\pi-\dfrac{4}{3}) \\=\dfrac{8}{9} (3\pi-4)$
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