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 960: 51

Answer

We prove the formula: ${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{1}{{\left| {\cos \phi } \right|}}{\rm{d}}A$

Work Step by Step

The graph $z = g\left( {x,y} \right)$ has the parametrization: $G\left( {x,y} \right) = \left( {x,y,g\left( {x,y} \right)} \right)$ So, ${{\bf{T}}_x} = \left( {1,0,{g_x}} \right)$ ${{\bf{T}}_y} = \left( {0,1,{g_y}} \right)$ Recall Theorem 1: ${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ where ${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y}$, a normal vector to the surface $S$. ${\bf{N}}\left( {x,y} \right) = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 1&0&{{g_x}}\\ 0&1&{{g_y}} \end{array}} \right| = - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}$ Let the $z$-axis be the vertical axis. Since $\phi = \phi \left( {x,y} \right)$ is the angle between the normal to $S$ and the vertical. So, ${\bf{N}}\left( {x,y} \right)\cdot{\bf{k}} = ||{\bf{N}}\left( {x,y} \right)||\cos \phi $ $\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right| = ||{\bf{N}}\left( {x,y} \right)||\left| {\cos \phi } \right|$ $||{\bf{N}}\left( {x,y} \right)|| = \dfrac{{\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right|}}{{\left| {\cos \phi } \right|}}$ Substituting $||{\bf{N}}\left( {x,y} \right)||$ into ${\rm{Area}}\left( S \right)$ gives ${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{{\left| {{\bf{N}}\left( {x,y} \right)\cdot{\bf{k}}} \right|}}{{\left| {\cos \phi } \right|}}{\rm{d}}x{\rm{d}}y$ But ${\bf{N}}\left( {x,y} \right)\cdot{\bf{k}} = \left( { - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}} \right)\cdot{\bf{k}} = 1$ and $dA = dxdy$. Therefore ${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{1}{{\left| {\cos \phi } \right|}}{\rm{d}}A$
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