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: 42

Answer

(a) We show that the circle generated is parametrized by $\left( {a\cos \theta ,a\sin \theta ,b} \right)$, ${\ \ \ \ \ }$ $0 \le \theta \le 2\pi $ (b) We show that $S$ is parametrized by $G\left( {y,\theta } \right) = \left( {y\cos \theta ,y\sin \theta ,g\left( y \right)} \right)$ for $c \le y \le d$, $0 \le \theta \le 2\pi $. (c) We prove the formula: $Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$

Work Step by Step

Refer to Figure 23. (a) Since the circle is generated by rotating about the $z$-axis, it is located on the plane that is parallel to the $xy$-plane. So, it has the equation $\left( {r\cos \theta ,r\sin \theta ,z} \right)$, ${\ \ \ \ \ }$ $0 \le \theta \le 2\pi $ Specifically, if we rotate a point $\left( {0,a,b} \right)$ about the $z$-axis, we have the radius of the circle to be $a$, and it is located on the plane $z=b$. Thus, we get the parametrization $\left( {a\cos \theta ,a\sin \theta ,b} \right)$, ${\ \ \ \ \ }$ $0 \le \theta \le 2\pi $ (b) From part (a) we get the parametrization of the circle: $\left( {a\cos \theta ,a\sin \theta ,b} \right)$, ${\ \ \ \ \ }$ $0 \le \theta \le 2\pi $ Since $S$ is formed by rotating the graph $z = g\left( y \right)$ about the $z$-axis as is shown in Figure 23, the radius is dependent on $y$; and the position of the circle is located on $z = g\left( y \right)$. Therefore, the parametrization of $S$ is $G\left( {y,\theta } \right) = \left( {y\cos \theta ,y\sin \theta ,g\left( y \right)} \right)$ for $c \le y \le d$, $0 \le \theta \le 2\pi $. (c) From part (b) we obtain Eq. (13): (13) ${\ \ \ \ \ }$ $G\left( {y,\theta } \right) = \left( {y\cos \theta ,y\sin \theta ,g\left( y \right)} \right)$ for $c \le y \le d$, $0 \le \theta \le 2\pi $. So, ${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {\cos \theta ,\sin \theta ,g'\left( y \right)} \right)$ ${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - y\sin \theta ,y\cos \theta ,0} \right)$ ${\bf{N}}\left( {y,\theta } \right) = {{\bf{T}}_y} \times {{\bf{T}}_\theta } = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ {\cos \theta }&{\sin \theta }&{g'\left( y \right)}\\ { - y\sin \theta }&{y\cos \theta }&0 \end{array}} \right|$ ${\bf{N}}\left( {y,\theta } \right) = - {\rm{yg}}'\left( {\rm{y}} \right)\cos {\rm{\theta }}{\bf{i}} - {\rm{yg}}'\left( {\rm{y}} \right)\sin {\rm{\theta }}{\bf{j}} + y{\bf{k}}$ $||{\bf{N}}\left( {y,\theta } \right)|| = \sqrt {{{\left( { - {\rm{yg}}'\left( {\rm{y}} \right)\cos {\rm{\theta }}} \right)}^2} + {{\left( { - y{\rm{g}}'\left( {\rm{y}} \right)\sin {\rm{\theta }}} \right)}^2} + {y^2}} $ $||{\bf{N}}\left( {y,\theta } \right)|| = \sqrt {{y^2}g'{{\left( y \right)}^2} + {y^2}} = y\sqrt {1 + g'{{\left( y \right)}^2}} $ By Theorem 1, the area of $S$ is $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {y,\theta } \right)||{\rm{d}}y{\rm{d}}\theta $ $Area\left( S \right) = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{y = c}^d ||{\bf{N}}\left( {y,\theta } \right)||{\rm{d}}y{\rm{d}}\theta $ $Area\left( S \right) = \mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta \mathop \smallint \limits_{y = c}^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$ Hence, $Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$.
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