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 902: 33

Answer

2

Work Step by Step

$\int^2_0 \int^{2-x}_0 \int^{4-2x-2y}_{(2-x-y)/2}$ dz dy dx =$\int^2_0 \int^{2-x}_0 (3-\frac{3x}{2}-\frac{3y}{2})dy $ dx =$\int^2_0 [3(1-\frac{x}{2})(2-x)-\frac{3}{4}(2-x)^2]dx $ =$\int^2_0 [6-6x+\frac{3x^2}{2}-\frac{3(2-x)}{4}]dx $ =$[6x-3x^2+\frac{x^3}{2}+\frac{(2-x)^3}{4}]^2_0$ =$12-12+4+0-\frac{2^3}{4}$ =2
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.