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: 33

Answer

(a) $Area\left( S \right) \approx 1.0781$ (b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S \approx 0.0981$

Work Step by Step

(a) The surface $z = \ln \left( {5 - {x^2} - {y^2}} \right)$ can be parametrized by $G\left( {x,y} \right) = \left( {x,y,\ln \left( {5 - {x^2} - {y^2}} \right)} \right)$. So, ${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0, - \dfrac{{2x}}{{5 - {x^2} - {y^2}}}} \right)$ ${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1, - \dfrac{{2y}}{{5 - {x^2} - {y^2}}}} \right)$ ${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 1&0&{ - \dfrac{{2x}}{{5 - {x^2} - {y^2}}}}\\ 0&1&{ - \dfrac{{2y}}{{5 - {x^2} - {y^2}}}} \end{array}} \right|$ $ = \dfrac{{2x}}{{5 - {x^2} - {y^2}}}{\bf{i}} + \dfrac{{2y}}{{5 - {x^2} - {y^2}}}{\bf{j}} + {\bf{k}}$ $||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( {\dfrac{{2x}}{{5 - {x^2} - {y^2}}}} \right)}^2} + {{\left( {\dfrac{{2y}}{{5 - {x^2} - {y^2}}}} \right)}^2} + 1} $ $ = \sqrt {\dfrac{{4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}}}{{{{\left( {5 - {x^2} - {y^2}} \right)}^2}}}} = \dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}$ By Theorem 1, the surface area of $S$ is $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ $Area\left( S \right) = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}{\rm{d}}y{\rm{d}}x$ Using a computer algebra system, we evaluate the integral and obtain $Area\left( S \right) \approx 1.0781$ (b) By Eq. (7) of Theorem 1, we have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ where $f\left( {x,y,z} \right) = {x^2}{y^3}$. We have $f\left( {G\left( {x,y} \right)} \right) = {x^2}{y^3}$. So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {{x^2}{y^3}} \right)\left( {\dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}} \right){\rm{d}}x{\rm{d}}y$ Using a computer algebra system, we evaluate the integral and obtain $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S \approx 0.0981$
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