Thomas' Calculus 13th Edition

Published by Pearson

Chapter 16: Integrals and Vector Fields - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 990: 44

Answer

$$\dfrac{\pi}{3}$$

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, $r^2=x^2+y^2+z^2$ $$Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} \space dA\\=(2) \times \space \int_{-1/2}^{1/2} \int_{0}^{1/2}\dfrac{1}{\sqrt {1-x^2} } \space dy \space dx \\=[\sin^{-1}x]_{-1/2}^{1/2} \\=\dfrac{\pi}{3}$$

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