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 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 875: 31



Work Step by Step

$\int^{2}_{1} \int^{3}_{1}kx^2y dx dy$ =$\int^{2}_{1}[\frac{k}{3}x^3y]^{x=3}_{x=0}]dx$ =$\int^{2}_{1}9ky dy= \frac{9}{2}ky^]| ^2_1$ =k=$\frac{2}{27}$ =Thus we choose k=$\frac{2}{27}$
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.