Answer
(a)
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0,y} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1,x} \right)$
${\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right) = \left( { - {\rm{y}}, - {\rm{x}}, - 1} \right)$
(b) We verify the following formula and evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {1 + {x^2} + {y^2}} {\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \frac{\pi }{6}\left( {2\sqrt 2 - 1} \right)$
(c) We verify the following formula and evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \mathop \smallint \limits_0^{\pi /2} \mathop \smallint \limits_0^1 \left( {\sin \theta \cos \theta } \right){r^3}\sqrt {1 + {r^2}} {\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{{15}}\left( {1 + \sqrt 2 } \right)$
Work Step by Step
(a) We have $G\left( {x,y} \right) = \left( {x,y,xy} \right)$. So,
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0,y} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1,x} \right)$
${\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&y\\
0&1&x
\end{array}} \right| = - y{\bf{i}} - x{\bf{j}} - {\bf{k}}$
So, ${\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right) = \left( { - {\rm{y}}, - {\rm{x}}, - 1} \right)$.
(b) From part (a) we obtain ${\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right) = \left( { - {\rm{y}}, - {\rm{x}}, - 1} \right)$. Hence,
$||{\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right)|| = \sqrt {{{\left( { - y} \right)}^2} + {{\left( { - x} \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {1 + {x^2} + {y^2}} $
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
Hence,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {1 + {x^2} + {y^2}} {\rm{d}}x{\rm{d}}y$
We compute this integral using polar coordinates by defining ${\cal D} = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1,x \ge 0,y \ge 0} \right\}$ in polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right):0 \le r \le 1,0 \le \theta \le \frac{\pi }{2}} \right\}$
Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{\pi /2} \mathop \smallint \limits_{r = 0}^1 \sqrt {1 + {r^2}} r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{\pi /2} {\rm{d}}\theta \mathop \smallint \limits_{r = 0}^1 \sqrt {1 + {r^2}} r{\rm{d}}r = \frac{\pi }{2}\mathop \smallint \limits_{r = 0}^1 \sqrt {1 + {r^2}} r{\rm{d}}r$
Let $t = 1 + {r^2}$. So, ${\rm{d}}t = 2r{\rm{d}}r$.
So,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = \frac{\pi }{4}\mathop \smallint \limits_{t = 1}^2 \sqrt t {\rm{d}}t = \frac{\pi }{4}\cdot\frac{2}{3}\left( {{t^{3/2}}|_1^2} \right) = \frac{\pi }{6}\left( {2\sqrt 2 - 1} \right)$
(c) By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} z||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
From $G\left( {x,y} \right) = \left( {x,y,xy} \right)$ we obtain $z = xy$. Since in polar coordinates: $x = r\cos \theta $ and $y = r\sin \theta $, so we get $z = {r^2}\cos \theta \sin \theta $.
Also, in polar coordinates $||{\bf{N}}\left( {{\rm{x}},{\rm{y}}} \right)|| = \sqrt {1 + {x^2} + {y^2}} $ becomes
$||{\bf{N}}\left( {r,{\rm{\theta }}} \right)|| = \sqrt {1 + {r^2}} $
Thus, in polar coordinates we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {r^2}\cos \theta \sin \theta \sqrt {1 + {r^2}} r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \mathop \smallint \limits_0^{\pi /2} \mathop \smallint \limits_0^1 \left( {\sin \theta \cos \theta } \right){r^3}\sqrt {1 + {r^2}} {\rm{d}}r{\rm{d}}\theta $
Next, we evaluate this integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \mathop \smallint \limits_0^{\pi /2} \sin \theta \cos \theta {\rm{d}}\theta \mathop \smallint \limits_0^1 {r^3}\sqrt {1 + {r^2}} {\rm{d}}r$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{2}\mathop \smallint \limits_0^{\pi /2} \sin 2\theta {\rm{d}}\theta \mathop \smallint \limits_0^1 {r^3}\sqrt {1 + {r^2}} {\rm{d}}r$
Since the derivative of $\frac{1}{{15}}\left( {3{r^2} - 2} \right){\left( {1 + {r^2}} \right)^{3/2}}$ is ${r^3}\sqrt {1 + {r^2}} $, we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = - \frac{1}{4}\left( {\cos 2\theta |_0^{\pi /2}} \right)\cdot\frac{1}{{15}}\left[ {\left( {3{r^2} - 2} \right){{\left( {1 + {r^2}} \right)}^{3/2}}} \right]_0^1$
$ = - \frac{1}{{60}}\left( { - 2} \right)\left( {2\sqrt 2 + 2} \right)$
$ = \frac{1}{{15}}\left( {\sqrt 2 + 1} \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} z{\rm{d}}S = \frac{1}{{15}}\left( {1 + \sqrt 2 } \right)$.
