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 901: 25



Work Step by Step

V=$\int^4_0 \int^{\sqrt{4-x}}_0 \int^{2-y}_0$ dz dy dx =$\int^4_0 \int^{\sqrt{4-x}}_0 (2-y)$ dy dx =$\int^4_[2\sqrt{4-x}-(\frac{4-x}{2})]dx $ =$[-\frac{4}{3}(4-x)^{3/2}+\frac{1}{4}(4-x^2)]^4_0$ =$\frac{4}{3}(4)^{3/2}-\frac{1}{4}(16)$ =$\frac{32}{3}-4$ =$\frac{20}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.