Answer
(a)
${{\bf{T}}_u} = \left( {2,1,3} \right)$
${{\bf{T}}_v} = \left( {0, - 1,1} \right)$
${\bf{N}}\left( {{\rm{u}},{\rm{v}}} \right) = 4{\bf{i}} - 2{\bf{j}} - 2{\bf{k}}$
(b) ${\rm{Area}}\left( S \right) = 4\sqrt 6 $
(c) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \dfrac{{32}}{3}\sqrt 6 $
Work Step by Step
From $G\left( {u,v} \right) = \left( {2u + 1,u - v,3u + v} \right)$, we have $x = 2u + 1$, $y = u - v$, and $z = 3u + v$. Since
$2x - y - z = 2\left( {2u + 1} \right) - \left( {u - v} \right) - \left( {3u + v} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $ = 4u + 2 - u + v - 3u - v$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $ = 2$
So, $G\left( {u,v} \right) = \left( {2u + 1,u - v,3u + v} \right)$ parametrizes the plane $2x - y - z = 2$.
(a)
${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2,1,3} \right)$
${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {0, - 1,1} \right)$
${\bf{N}}\left( {{\rm{u}},{\rm{v}}} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&1&3\\
0&{ - 1}&1
\end{array}} \right| = 4{\bf{i}} - 2{\bf{j}} - 2{\bf{k}}$
(b) From the result in part (a) we get ${\bf{N}}\left( {{\rm{u}},{\rm{v}}} \right) = \left( {4, - 2, - 2} \right)$. So,
$||{\bf{N}}\left( {u,v} \right)|| = \sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt {24} = 2\sqrt 6 $
Next, we compute the area of $S = G\left( {\cal D} \right)$, where ${\cal D} = \left\{ {\left( {u,v} \right):0 \le u \le 2,0 \le v \le 1} \right\}$:
${\rm{Area}}\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$
${\rm{Area}}\left( S \right) = 2\sqrt 6 \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = 0}^2 {\rm{d}}u{\rm{d}}v = 2\sqrt 6 \cdot2 = 4\sqrt 6 $
(c) From $G\left( {u,v} \right) = \left( {2u + 1,u - v,3u + v} \right)$, we have $x = 2u + 1$, $y = u - v$, and $z = 3u + v$.
1. Express $f\left( {x,y,z} \right) = yz$ in terms of $u$ and $v$
$f\left( {u,v} \right) = \left( {u - v} \right)\left( {3u + v} \right) = 3{u^2} - 2uv - {v^2}$
2. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S$
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {u,v} \right)} \right)||{\bf{N}}\left( {u,v} \right)||{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 2\sqrt 6 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {u,v} \right)} \right){\rm{d}}u{\rm{d}}v$
$ = 2\sqrt 6 \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = 0}^2 \left( {3{u^2} - 2uv - {v^2}} \right){\rm{d}}u{\rm{d}}v$
$ = 2\sqrt 6 \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = 0}^2 3{u^2}{\rm{d}}u{\rm{d}}v - 4\sqrt 6 \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = 0}^2 uv{\rm{d}}u{\rm{d}}v - 2\sqrt 6 \mathop \smallint \limits_{v = 0}^1 \mathop \smallint \limits_{u = 0}^2 {v^2}{\rm{d}}u{\rm{d}}v$
$ = 2\sqrt 6 \left( {{u^3}|_0^2} \right)\left( {v|_0^1} \right) - 4\sqrt 6 \left( {\frac{1}{2}{u^2}|_0^2} \right)\left( {\frac{1}{2}{v^2}|_0^1} \right) - 2\sqrt 6 \left( {u|_0^2} \right)\left( {\frac{1}{3}{v^3}|_0^1} \right)$
$ = 16\sqrt 6 - 4\sqrt 6 - \frac{4}{3}\sqrt 6 = \frac{{32}}{3}\sqrt 6 $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \frac{{32}}{3}\sqrt 6 $.
