Calculus (3rd Edition)

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

Chapter 17 - Line and Surface Integrals - 17.5 Surface Integrals of Vector Fields - Exercises - Page 970: 38

Answer

Using a computer algebra system (a) we verify that $||{\bf{N}}\left( {u,v} \right)|{|^2} = 1 + \frac{3}{4}{v^2} + 2v\cos \frac{u}{2} + \frac{1}{2}{v^2}\cos u$ (b) $Area\left( S \right) \approx 6.3533$ (c) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S \approx 7.4003$

Work Step by Step

(a) We have $G\left( {u,v} \right) = \left( {\left( {1 + v\cos \frac{u}{2}} \right)\cos u,\left( {1 + v\cos \frac{u}{2}} \right)\sin u,v\sin \frac{u}{2}} \right)$ for $0 \le u \le 2\pi $, $ - \frac{1}{2} \le v \le \frac{1}{2}$. Using a computer algebra system, we obtain ${{\bf{T}}_u} = ( - \frac{1}{2}v\cos u\sin \frac{u}{2} - \left( {1 + v\cos \frac{u}{2}} \right)\sin u,$ $ - \frac{1}{2}v\sin u\sin \frac{u}{2} + \left( {1 + v\cos \frac{u}{2}} \right)\cos u,\frac{1}{2}v\cos \frac{u}{2})$ ${{\bf{T}}_v} = \left( {\cos \frac{u}{2}\cos u,\cos \frac{u}{2}\sin u,\sin \frac{u}{2}} \right)$ ${\bf{N}}\left( {u,v} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = (\frac{1}{2}\left( { - v\cos \frac{u}{2} + 2\cos u + v\cos \frac{{3u}}{2}} \right)\sin \frac{u}{2},$ $\frac{1}{4}\left( {v + 2\cos \frac{u}{2} + 2v\cos u - 2\cos \frac{{3u}}{2} - v\cos 2u} \right),$ $ - \cos \frac{u}{2}\left( {1 + v\cos \frac{u}{2}} \right))$ Since $||{\bf{N}}\left( {u,v} \right){|^2} = {\bf{N}}\left( {u,v} \right)\cdot{\bf{N}}\left( {u,v} \right)$, we compute this equation using a computer algebra system. The results is $||{\bf{N}}\left( {u,v} \right)|{|^2} = 1 + \frac{3}{4}{v^2} + 2v\cos \frac{u}{2} + \frac{1}{2}{v^2}\cos u$ (b) From part (a), we obtain $||{\bf{N}}\left( {u,v} \right)|{|^2} = 1 + \frac{3}{4}{v^2} + 2v\cos \frac{u}{2} + \frac{1}{2}{v^2}\cos u$ To compute the surface area, we use the following equation (Section 17.4) $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$ $Area\left( S \right) = \mathop \smallint \limits_{v = - \frac{1}{2}}^{\frac{1}{2}} \mathop \smallint \limits_{u = 0}^{2\pi } \sqrt {1 + \frac{3}{4}{v^2} + 2v\cos \frac{u}{2} + \frac{1}{2}{v^2}\cos u} {\rm{d}}u{\rm{d}}v$ Using a computer algebra system, the result is $Area\left( S \right) \approx 6.3533$ (c) To compute $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S$, we write $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$. By Eq. (7) in Section 17.4: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {u,v} \right)} \right)||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$ Since $G\left( {u,v} \right) = \left( {\left( {1 + v\cos \frac{u}{2}} \right)\cos u,\left( {1 + v\cos \frac{u}{2}} \right)\sin u,v\sin \frac{u}{2}} \right)$, so $f\left( {G\left( {u,v} \right)} \right) = {\left( {1 + v\cos \frac{u}{2}} \right)^2}{\cos ^2}u + {\left( {1 + v\cos \frac{u}{2}} \right)^2}{\sin ^2}u + {v^2}{\sin ^2}\frac{u}{2}$ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S$ $ = \mathop \smallint \limits_{v = - \frac{1}{2}}^{\frac{1}{2}} \mathop \smallint \limits_{u = 0}^{2\pi } \left[ {{{\left( {1 + v\cos \frac{u}{2}} \right)}^2}{{\cos }^2}u + {{\left( {1 + v\cos \frac{u}{2}} \right)}^2}{{\sin }^2}u + {v^2}{{\sin }^2}\frac{u}{2}} \right]$ ${\ \ \ \ }$ $\left( {\sqrt {1 + \frac{3}{4}{v^2} + 2v\cos \frac{u}{2} + \frac{1}{2}{v^2}\cos u} } \right){\rm{d}}u{\rm{d}}v$ Using a computer algebra system, this yields $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}S \approx 7.4003$
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