Thomas' Calculus 13th Edition

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

Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 25


$$\sqrt 6$$

Work Step by Step

We have $ r_u=1+j \space and \\ r_v=1-j+k $ and $$|r_u \times r_v|=\sqrt 6$$ $$ Surface \space Area=\iint_S |r_u \times r_v| \space du \space dv \\=\int_0^1 \int_0^1 \sqrt 6 \space du \space dv \\=\sqrt 6(1-0) \\=\sqrt 6$$
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.