Answer
(a) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = 0$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \pi $
(c) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \pi $
Work Step by Step
Since $S$ is oriented with normal in the positive $z$-direction, we have ${\bf{n}} = \left( {0,0,1} \right)$.
(a)
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left[ {\left( {1,0,0} \right)\cdot\left( {0,0,1} \right)} \right]{\rm{d}}S = 0$
(b)
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left[ {\left( {0,0,1} \right)\cdot\left( {0,0,1} \right)} \right]{\rm{d}}S$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = Area\left( S \right)$
Since the area of the disk ${x^2} + {y^2} \le 1$ is $\pi $, so $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \pi $.
(c)
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left[ {\left( {1,1,1} \right)\cdot\left( {0,0,1} \right)} \right]{\rm{d}}S$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S = Area\left( S \right)$
Since the area of the disk ${x^2} + {y^2} \le 1$ is $\pi $, so $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \pi $.
