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 - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 990: 49


$$\dfrac{(13\sqrt {13}-1) \pi}{6} $$

Work Step by Step

Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$ We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$ $ r_r=\lt \cos \theta,\sin \theta, 2r \gt ; r_{\theta}=\lt -r\sin \theta, r\cos \theta,0 \gt $ Also, $|r_r \times r_{\theta}|=r\sqrt {4r^2+1}$ Now, $ Area=\int_0^{2 \pi} \int_1^{\sqrt 3} (r\sqrt {4r^2+1}) \space dr \space d \theta \\=\int_0^{2 \pi} (\dfrac{13\sqrt {13}}{12}-1) \space d \theta \\=\dfrac{(13\sqrt {13}-1) \pi}{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.