Answer
(a) zero line integral
(b) nonzero
(c) zero line integral
(d) zero line integral
(e) zero line integral
(f) nonzero
Work Step by Step
The vertical segment $C$ from $\left( {0,0} \right)$ to $\left( {0,1} \right)$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {0,t} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
So, $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {0,1} \right)\cdot\left( {0,1} \right)} = 1$.
(a) We have $f\left( {x,y} \right) = x$. Since $x=0$ from $\left( {0,0} \right)$ to $\left( {0,1} \right)$, so $f\left( {x,y} \right) = 0$.
Therefore, $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = 0$
So, $f\left( {x,y} \right) = x$ has a zero line integral over the vertical segment from $\left( {0,0} \right)$ to $\left( {0,1} \right)$.
(b) We have $f\left( {x,y} \right) = y$.
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$ = \mathop \smallint \limits_0^1 t{\rm{d}}t = \frac{1}{2}{t^2}|_0^1 = \frac{1}{2}$
So, $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \frac{1}{2} \ne 0$.
(c) We have ${\bf{F}} = \left( {x,0} \right)$.
By Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
Since ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {0,0} \right)$, so $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$.
So, ${\bf{F}} = \left( {x,0} \right)$ has a zero line integral over the vertical segment from $\left( {0,0} \right)$ to $\left( {0,1} \right)$.
(d) We have ${\bf{F}} = \left( {y,0} \right)$.
By Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {t,0} \right)\cdot\left( {0,1} \right){\rm{d}}t = 0$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$.
So, ${\bf{F}} = \left( {y,0} \right)$ has a zero line integral over the vertical segment from $\left( {0,0} \right)$ to $\left( {0,1} \right)$.
(e) We have ${\bf{F}} = \left( {0,x} \right)$.
By Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
Since ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {0,0} \right)$, so $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$.
So, ${\bf{F}} = \left( {0,x} \right)$ has a zero line integral over the vertical segment from $\left( {0,0} \right)$ to $\left( {0,1} \right)$.
(f) We have ${\bf{F}} = \left( {0,y} \right)$.
By Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {0,t} \right)\cdot\left( {0,1} \right){\rm{d}}t = \mathop \smallint \limits_0^1 t{\rm{d}}t$
$ = \frac{1}{2}{t^2}|_0^1 = \frac{1}{2}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{2} \ne 0$.
