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: 13

Answer

18

Work Step by Step

$\int^3_0 \int^{\sqrt{9-x^2}}_0 \int^{\sqrt{9-x^2}}_0 $ dz dy dx =$\int^3_0 \int^{\sqrt{9-x^2}}_0 \sqrt{9-x^2}$ dy dx =$\int^3_0 (9-x^2)$ dy dx =$\int^3_0 (9-x^2)$ dx =$[9x-\frac{x^3}{3}]^3_0$ =18
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