Answer
(a) $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
(b) $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 5$
Work Step by Step
(a) If ${\bf{F}}\left( P \right)$ is normal to $C$ at all points $P$ on $C$, then ${\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$ along the curve $C$. Therefore,
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
(b) If ${\bf{F}}\left( P \right)$ is a unit vector pointing in the negative direction along the curve, then Eq. (7) becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {{\bf{F}}\cdot{\bf{T}}} \right){\rm{d}}s = - \mathop \smallint \limits_C^{} {\rm{d}}s$
where ${\bf{T}}$ is the unit tangent vector on $C$.
Since $C$ has length 5, so $\mathop \smallint \limits_C^{} {\rm{d}}s = 5$. Therefore,
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 5$
