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: 36

Answer

$\frac{4}{7}$

Work Step by Step

$ V=2\int^1_0 \int^{1-y^2}_0 \int^{x^2+y^2}_0$ dz dx dy =$2\int^2_0 \int^{1-y^2}_0 (x^2+y^2)dx $ dy =$2\int^1_0 [\frac{x^3}{3}+xy^2]^{1-y^2}_0$ dy =$2\int^1_0 (1-y^2)[\frac{1}{3}(1-y^2)+y^2]dy $ =$2\int^1_0 (1-y^2)(\frac{1}{3}+\frac{1}{3}y^2+\frac{1}{3}y^4)dy $ =$\frac{2}{3}\int^1_0 (1-y^6)dy $ =$\frac{2}{3}[y-\frac{y^7}{7}]$ =$(\frac{2}{3})(\frac{6}{7})$ =$\frac{4}{7}$
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.