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: 34

Answer

$Area\left( S \right) = \frac{3}{2}\sqrt {29} $

Work Step by Step

From the equation of the plane $2x + 3y + 4z = 28$, we can write $z = \dfrac{1}{4}\left( {28 - 2x - 3y} \right)$. So, the plane can be parametrized by $G\left( {x,y} \right) = \left( {x,y,\dfrac{1}{4}\left( {28 - 2x - 3y} \right)} \right)$. ${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0, - \dfrac{1}{2}} \right)$ ${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1, - \dfrac{3}{4}} \right)$ ${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 1&0&{ - \dfrac{1}{2}}\\ 0&1&{ - \dfrac{3}{4}} \end{array}} \right| = \dfrac{1}{2}{\bf{i}} + \dfrac{3}{4}{\bf{j}} + {\bf{k}}$ $||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{3}{4}} \right)}^2} + 1} = \dfrac{{\sqrt {29} }}{4}$ 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) = \dfrac{{\sqrt {29} }}{4}\mathop \smallint \limits_{x = 1}^3 \mathop \smallint \limits_{y = 2}^5 {\rm{d}}y{\rm{d}}x = \dfrac{{\sqrt {29} }}{4}\left( {x|_1^3} \right)\left( {y|_2^5} \right) = \dfrac{{\sqrt {29} }}{4}\cdot2\cdot3 = \dfrac{3}{2}\sqrt {29} $
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