Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S \approx 0.6$
Work Step by Step
We can approximate the integral by
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S \approx \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_1}}^{} 0.9{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_2}}^{} 1{\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_3}}^{} 1.1{\rm{d}}S$
$ = 0.9\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_1}}^{} {\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_2}}^{} {\rm{d}}S + 1.1\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_3}}^{} {\rm{d}}S$
$ = 0.9 \times 0.2 + 0.2 + 1.1 \times 0.2$
$ = 0.6$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S \approx 0.6$.
