Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 12



Work Step by Step

We calculate the integral as follows: $$I=\int_{0}^{1} \int_{ 3 \sqrt y} \dfrac{2 \pi \sin (\pi x^2)}{x^2} \ dx \ dy \\= 2 \times \int_{0}^{1} \int_{0}^{x^3} \dfrac{ \pi \sin (\pi x^2)}{x^2} \ dy \ dx \\=2 \int_{0}^{1} \pi \times x \times \sin (\pi x^2) \ dx \\=-[\cos (\pi x^2) ]_0^1 \\=-(-1)-(-1) \\=2$$
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.