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: 36

Answer

$Area\left( S \right) = 21\sqrt 2 \pi $

Work Step by Step

Since the surface of the part of the cone ${x^2} + {y^2} = {z^2}$ is located between the planes $z=2$ and $z=5$, we have the domain ${\cal D}$ given by ${\cal D} = \left\{ {\left( {x,y} \right):4 \le {x^2} + {y^2} \le 25} \right\}$ We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y, \pm \sqrt {{x^2} + {y^2}} } \right)$. Since $z$ is positive, we choose the parametrization $G\left( {x,y} \right) = \left( {x,y,\sqrt {{x^2} + {y^2}} } \right)$. ${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0,\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$ ${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1,\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)$ ${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 1&0&{\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}}\\ 0&1&{\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \end{array}} \right|$ ${\bf{N}}\left( {x,y} \right) = - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}} + {\bf{k}}$ $||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( { - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + {{\left( { - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + 1} $ $||{\bf{N}}\left( {x,y} \right)|| = \sqrt {\dfrac{{2{x^2} + 2{y^2}}}{{{x^2} + {y^2}}}} = \sqrt 2 $ By Theorem 1, the surface area of $S$ is $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ $Area\left( S \right) = \sqrt 2 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$ Notice that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$ is the area of the annular region ${\cal D} = \left\{ {\left( {x,y} \right):4 \le {x^2} + {y^2} \le 25} \right\}$. So, we have $Area\left( {\cal D} \right) = \pi \cdot{5^2} - \pi \cdot{2^2} = 21\pi $ Thus, the surface of the part of the cone is $Area\left( S \right) = 21\sqrt 2 \pi $
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