Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \dfrac{{{x^2}}}{{4 - z}}{\rm{d}}S = \dfrac{\pi }{{12}}\left( {17\sqrt {17} - 5\sqrt 5 } \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = \dfrac{{{x^2}}}{{4 - z}}$.
We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y,4 - {x^2} - {y^2}} \right)$. So,
${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0, - 2x} \right)$
${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1, - 2y} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - 2x}\\
0&1&{ - 2y}
\end{array}} \right|$
${\bf{N}}\left( {x,y} \right) = 2x{\bf{i}} + 2y{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,z} \right)|| = \sqrt {{{\left( {2x} \right)}^2} + {{\left( {2y} \right)}^2} + {1^2}} = \sqrt {1 + 4\left( {{x^2} + {y^2}} \right)} $
Since $z = 4 - {x^2} - {y^2}$, so $f\left( {G\left( {x,y} \right)} \right) = \dfrac{{{x^2}}}{{{x^2} + {y^2}}}$.
From $z = 4 - {x^2} - {y^2}$ and $0 \le z \le 3$ we obtain
$0 \le 4 - {x^2} - {y^2} \le 3$
$ - 4 \le - {x^2} - {y^2} \le - 1$
$1 \le {x^2} + {y^2} \le 4$
From here we obtain the description of the domain:
${\cal D} = \left\{ {\left( {x,y} \right):1 \le {x^2} + {y^2} \le 4} \right\}$
In polar coordinates: ${\cal D} = \left\{ {\left( {r,\theta } \right):1 \le r \le 2,0 \le \theta \le 2\pi } \right\}$.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \dfrac{{{x^2}}}{{{x^2} + {y^2}}}\sqrt {1 + 4\left( {{x^2} + {y^2}} \right)} {\rm{d}}x{\rm{d}}y$
We evaluate this integral using polar coordinates: $x = r\cos \theta $ and $y = r\sin \theta $.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 1}^2 {\cos ^2}\theta \sqrt {1 + 4{r^2}} r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\cos }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 1}^2 \sqrt {1 + 4{r^2}} r{\rm{d}}r} \right)$
Since $\cos 2\theta = 2{\cos ^2}\theta - 1$, so ${\cos ^2}\theta = \dfrac{{\cos 2\theta + 1}}{2}$. The integral above becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \dfrac{{\cos 2\theta + 1}}{2}{\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 1}^2 \sqrt {1 + 4{r^2}} r{\rm{d}}r} \right)$
Let $t = 1 + 4{r^2}$. So, ${\rm{d}}t = 8r{\rm{d}}r$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\dfrac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta + \dfrac{1}{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } \cos 2\theta {\rm{d}}\theta } \right)\left( {\dfrac{1}{8}\mathop \smallint \limits_{r = 5}^{17} \sqrt t {\rm{d}}t} \right)$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \left( {\pi + \dfrac{1}{4}\sin 2\theta |_0^{2\pi }} \right)\left( {\dfrac{1}{8}\cdot\dfrac{2}{3}{t^{3/2}}|_5^{17}} \right) = \dfrac{\pi }{{12}}\left( {17\sqrt {17} - 5\sqrt 5 } \right)$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \dfrac{{{x^2}}}{{4 - z}}{\rm{d}}S = \dfrac{\pi }{{12}}\left( {17\sqrt {17} - 5\sqrt 5 } \right)$.
