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: 37

Answer

$Area\left( S \right) = \pi $

Work Step by Step

From ${z^2} = {x^2} + {y^2}$ and ${y^2} + {z^2} \le 1$ we get ${y^2} + {z^2} = {x^2} + 2{y^2} \le 1$ So, we can consider the domain ${\cal D}$ to be described by ${\cal D} = \left\{ {\left( {x,y} \right):{x^2} + 2{y^2} \le 1} \right\}$ We notice that the boundary of ${\cal D}$ is an ellipse given by ${x^2} + {\left( {\dfrac{y}{{1/\sqrt 2 }}} \right)^2} = 1$, where the semimajor axis and the semiminor axis are $a=1$ and $b = \dfrac{1}{{\sqrt 2 }}$, respectively. We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y, \pm \sqrt {{x^2} + {y^2}} } \right)$. Since $z \ge 0$, 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 ellipse ${\cal D} = \left\{ {\left( {x,y} \right):{x^2} + 2{y^2} \le 1} \right\}$, where the semimajor axis and the semiminor axis are $a=1$ and $b = \dfrac{1}{{\sqrt 2 }}$, respectively. In Exercise 24 of Section 14.6 on page 753, we know that the area of the ellipse is $A = \pi ab$. So, $Area\left( S \right) = \sqrt 2 \cdot \pi \cdot 1 \cdot \dfrac{1}{{\sqrt 2 }} = \pi $
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