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 958: 12

Answer

Please see the figure attached. $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx 0.0424$

Work Step by Step

The small patch (green) is illustrated in the figure attached. From Exercise 9, we get $G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$. So, ${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - \sin \theta \sin \phi ,\cos \theta \sin \phi ,0} \right)$ ${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {\cos \theta \cos \phi ,\sin \theta \cos \phi , - \sin \phi } \right)$ ${\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right) = {{\bf{T}}_\theta } \times {{\bf{T}}_\phi } = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ { - \frac{1}{2}\sqrt 2 }&0&0\\ 0&{\frac{1}{2}\sqrt 2 }&{ - \frac{1}{2}\sqrt 2 } \end{array}} \right|$ ${\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right) = - \frac{1}{2}{\bf{j}} - \frac{1}{2}{\bf{k}}$ $||{\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right)|| = \sqrt {0 + {{\left( { - \frac{1}{2}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} = \frac{1}{2}\sqrt 2 $ We estimate the area of the small patch $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$ by $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx ||{\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right)||\Delta \theta \Delta \phi $ Since $\frac{\pi }{2} - 0.15 \le \theta \le \frac{\pi }{2} + 0.15$ and $\frac{\pi }{4} - 0.1 \le \phi \le \frac{\pi }{4} + 0.1$, so $\Delta \theta = 0.3$ and $\Delta \phi = 0.2$. $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx \frac{1}{2}\sqrt 2 \cdot 0.3 \cdot 0.2 = 0.0424$
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