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 989: 20


$$\dfrac{7 \pi \space \sqrt {10}}{3}$$

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$ We have $ z=1; z=\dfrac{4}{3} \implies z=\dfrac{\sqrt {x^2+y^2}}{3}$ Now, $ r_r= \cos \theta \space i+ \sin \theta \space j+\dfrac{1}{3} \space k\\ r_{\theta}=-r \sin \theta \space i+(r \cos \theta) j $ $|r_r \times r_{\theta}|=\dfrac{r \sqrt {10}}{3}$ Now, $$ Area=\int_0^{2 \pi} \int_3^{4} \dfrac{r \sqrt {10}}{3} \space dr \space d \theta \\=\int_0^{2 \pi} \dfrac{7\sqrt {10}}{6} \space d \theta \\=\dfrac{7 \pi \space \sqrt {10}}{3}$$
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