Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} y{\rm{d}}S = 2\pi $
Work Step by Step
Since $0 \le y \le 1$, so $y$ is always positive. We can parametrize the sphere ${x^2} + {y^2} + {z^2} = 4$ by $G\left( {x,z} \right) = \left( {x,\sqrt {4 - {x^2} - {z^2}} ,z} \right)$.
${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1, - \dfrac{x}{{\sqrt {4 - {x^2} - {z^2}} }},0} \right)$
${{\bf{T}}_z} = \dfrac{{\partial G}}{{\partial z}} = \left( {0, - \dfrac{z}{{\sqrt {4 - {x^2} - {z^2}} }},1} \right)$
${\bf{N}}\left( {x,z} \right) = {{\bf{T}}_x} \times {{\bf{T}}_z} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&{ - \dfrac{x}{{\sqrt {4 - {x^2} - {z^2}} }}}&0\\
0&{ - \dfrac{z}{{\sqrt {4 - {x^2} - {z^2}} }}}&1
\end{array}} \right|$
${\bf{N}}\left( {x,z} \right) = - \dfrac{x}{{\sqrt {4 - {x^2} - {z^2}} }}{\bf{i}} - {\bf{j}} - \dfrac{z}{{\sqrt {4 - {x^2} - {z^2}} }}{\bf{k}}$
$||{\bf{N}}\left( {x,z} \right)|| = \sqrt {\dfrac{{{x^2}}}{{4 - {x^2} - {z^2}}} + 1 + \dfrac{{{z^2}}}{{4 - {x^2} - {z^2}}}} = \sqrt {\dfrac{{4 - {x^2} - {z^2} + {x^2} + {z^2}}}{{4 - {x^2} - {z^2}}}} $
$||{\bf{N}}\left( {x,z} \right)|| = \dfrac{2}{{\sqrt {4 - {x^2} - {z^2}} }}$
From ${x^2} + {y^2} + {z^2} = 4$ and $0 \le y \le 1$, we obtain
$0 \le 4 - {x^2} - {z^2} \le 1$
$ - 4 \le - {x^2} - {z^2} \le - 3$
$3 \le {x^2} + {z^2} \le 4$
So, the domain description is ${\cal D} = \left\{ {\left( {x,z} \right):3 \le {x^2} + {z^2} \le 4} \right\}$.
In polar coordinates:
${\cal D} = \left\{ {\left( {r,\theta } \right):\sqrt 3 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Since $f\left( {x,y,z} \right) = y$, we obtain $f\left( {G\left( {x,z} \right)} \right) = \sqrt {4 - {x^2} - {z^2}} $.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,z} \right)} \right)||{\bf{N}}\left( {x,z} \right)||{\rm{d}}x{\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {4 - {x^2} - {z^2}} \left( {\dfrac{2}{{\sqrt {4 - {x^2} - {z^2}} }}} \right){\rm{d}}x{\rm{d}}z$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 2\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}z$
We evaluate this integral using the polar coordinates in the $xz$-plane with the domain description:
${\cal D} = \left\{ {\left( {r,\theta } \right):\sqrt 3 \le r \le 2,0 \le \theta \le 2\pi } \right\}$
Thus,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 2\mathop \smallint \limits_\theta ^{2\pi } \mathop \smallint \limits_{r = \sqrt 3 }^2 r{\rm{d}}r{\rm{d}}\theta $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = 2\left( {2\pi } \right)\left( {\dfrac{1}{2}{r^2}|_{\sqrt 3 }^2} \right) = 4\pi \left( {2 - \dfrac{3}{2}} \right) = 2\pi $
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} y{\rm{d}}S = 2\pi $.