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: 20

Answer

$8 \ln 2$

Work Step by Step

$\int^7_0 \int^2_0 \int^{\sqrt{4-q^2}}_0 \frac{q}{r+1}$ dp dq dr =$\int^7_0 \int^2_0 \frac{q\sqrt{4-q^2}}{r+1}$ dq dr =$\int^7_0 \frac{1}{3(r+1)}[-(4-q^2)^{3/2}]^2_0$ dr =$\frac{8}{3} \int^7_0 \frac{1}{r+1}dr $ =$\frac{8 \ln 8}{3}$ =$8 \ln 2$
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.