Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 959: 43

Answer

$Area\left( S \right) \simeq 36.177$

Work Step by Step

Using Eq. (14) we compute the surface area of $z = 4 - {y^2}$ for $0 \le y \le 2$ rotated about the $z$-axis: $Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$ $Area\left( S \right) = 2\pi \mathop \smallint \limits_0^2 y\sqrt {1 + {{\left( { - 2y} \right)}^2}} {\rm{d}}y = 2\pi \mathop \smallint \limits_0^2 y\sqrt {1 + 4{y^2}} {\rm{d}}y$ Write $t = 1 + 4{y^2}$. So, $dt = 8ydy$. The integral becomes $Area\left( S \right) = \frac{\pi }{4}\mathop \smallint \limits_1^{17} \sqrt t {\rm{d}}t$ $Area\left( S \right) = \frac{\pi }{4}\left( {\frac{2}{3}{t^{3/2}}|_1^{17}} \right) = \frac{\pi }{6}\left( {17\sqrt {17} - 1} \right) \simeq 36.177$
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