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 957: 4

Answer

(a) ${\rm{Area}}\left( S \right) = \frac{\pi }{2}\sqrt 6 $ (b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x - y} \right){\rm{d}}S = 2\sqrt 6 \left( {\frac{2}{3} + \frac{\pi }{4}} \right)$

Work Step by Step

(a) Using $G$ as defined in Exercise 3, that is, $G\left( {u,v} \right) = \left( {2u + 1,u - v,3u + v} \right)$, we compute ${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2,1,3} \right)$ ${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {0, - 1,1} \right)$ ${\bf{N}}\left( {{\rm{u}},{\rm{v}}} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 2&1&3\\ 0&{ - 1}&1 \end{array}} \right| = 4{\bf{i}} - 2{\bf{j}} - 2{\bf{k}}$ So, $||{\bf{N}}\left( {u,v} \right)|| = \sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt {24} = 2\sqrt 6 $ The surface area of S: ${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v = 2\sqrt 6 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}u{\rm{d}}v$ We compute this integral using polar coordinates by defining ${\cal D} = \left\{ {\left( {u,v} \right):{u^2} + {v^2} \le 1,u \ge 0,v \ge 0} \right\}$ in polar coordinates: ${\cal D} = \left\{ {\left( {r,\theta } \right):0 \le r \le 1,0 \le \theta \le \frac{\pi }{2}} \right\}$ Thus, the surface area becomes ${\rm{Area}}\left( S \right) = 2\sqrt 6 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}u{\rm{d}}v = 2\sqrt 6 \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} r{\rm{d}}r{\rm{d}}\theta $ ${\rm{Area}}\left( S \right) = 2\sqrt 6 \left( {\frac{\pi }{2}} \right)\left( {\frac{1}{2}{r^2}|_0^1} \right) = \frac{\pi }{2}\sqrt 6 $ (b) Since $G\left( {u,v} \right) = \left( {2u + 1,u - v,3u + v} \right)$, so $x - y = 2u + 1 - u + v = u + v + 1$ By Eq. (7) of Theorem 1, we have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {u,v} \right)} \right)||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x - y} \right){\rm{d}}S = 2\sqrt 6 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {u + v + 1} \right){\rm{d}}u{\rm{d}}v$ Similar to part (a), we evaluate this integral using polar coordinates, where $u = r\cos \theta $ and $v = r\sin \theta $. So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x - y} \right){\rm{d}}S = 2\sqrt 6 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {u + v + 1} \right){\rm{d}}u{\rm{d}}v$ $ = 2\sqrt 6 \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {r\cos \theta + r\sin \theta + 1} \right)r{\rm{d}}r{\rm{d}}\theta $ $ = 2\sqrt 6 \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} \left( {r\cos \theta + r\sin \theta + 1} \right)r{\rm{d}}r{\rm{d}}\theta $ $ = 2\sqrt 6 \left[ {\mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta + \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\theta = 0}^{\pi /2} r{\rm{d}}r{\rm{d}}\theta } \right]$ $ = 2\sqrt 6 \left[ {\frac{1}{3}\left( {{r^3}|_0^1} \right)\left( {\sin \theta |_0^{\pi /2}} \right) - \frac{1}{3}\left( {{r^3}|_0^1} \right)\left( {\cos \theta |_0^{\pi /2}} \right) + \frac{1}{2}\left( {{r^2}|_0^1} \right)\left( {\theta |_0^{\pi /2}} \right)} \right]$ $ = 2\sqrt 6 \left( {\frac{1}{3} + \frac{1}{3} + \frac{\pi }{4}} \right) = 2\sqrt 6 \left( {\frac{2}{3} + \frac{\pi }{4}} \right)$ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {x - y} \right){\rm{d}}S = 2\sqrt 6 \left( {\frac{2}{3} + \frac{\pi }{4}} \right)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.