Thomas' Calculus 13th Edition

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

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 14



Work Step by Step

We calculate the integral as follows: $$ Area=\int_{1}^{4} \int_{2-y}^{\sqrt y} \ dx \ dy \\=\int_{1}^{4} (y^{1/2}-2+y) \ dy \\=\int_{1}^{4} \sqrt y dy - \int_{1}^{4} 2 \ dy + \int_{1}^{4} y \ dy \\=[ \dfrac{2}{3} y^{3/2}]_{1}^{4} - 2[y]_{1}^{4}+[\dfrac{1}{2} \times y^2]_{1}^4 \\=\dfrac{14}{3}-6+\dfrac{15}{2} \\=\dfrac{37}{6}$$
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