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 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 900: 14

Answer

$\frac{16}{3}$

Work Step by Step

$\int^2_0 \int^{\sqrt{4-y^2}}_{\sqrt{4-y^2}} \int^{2x+y}_0$ dz dx dy =$\int^2_0 \int^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}(2x+y)$ dx dy =$\int^2_0[x^2+xy]^{\sqrt{4-y^2}}_{-{\sqrt{4-y^2}}}dy $ =$\int^2_0 (4-y^2)^{1/2}(2y)dy $ =$[-\frac{2}{3}(4-y^2)^{3/2}]^2_0$ =$\frac{2}{3}(4)^{3/2}$ =$\frac{16}{3}$
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