## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 16: Integrals and Vector Fields - Practice Exercises - Page 1029: 27

#### Answer

$$\pi[ \sqrt 2 +\ln (1+\sqrt 2)]$$

#### 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, $r^2=x^2+y^2+z^2$ $r_r=(\cos \theta) i+\sin \theta j; r_{\theta}=(-r\sin \theta) i+(r \cos \theta) j+k$ Now, $|r_r \times r_{\theta}|=\sqrt {1+r^2}$ $$Surface \space Area =\iint_S |r_u \times r_v| du dv \\=\int_0^{2 \pi} \int_0^1 \sqrt {1+r^2} dr d\theta \\ = \int_0^{2 \pi} \dfrac{\sqrt 2}{2}+\dfrac{\ln (1+\sqrt 2)}{2} ] d\theta \\=\pi[ \sqrt 2 +\ln (1+\sqrt 2)]$$

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