Answer
- the line integrals of $f$ along $ABC$ is
$\mathop \smallint \limits_{ABC}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq 7.6$
- the line integral of ${\bf{F}}$ along $ABC$ is
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq 4$
Work Step by Step
1. compute the line integral of $f$ along $ABC$
The line integral of $f$ along $ABC$ is estimated by the sum:
$\mathop \sum \limits_{i = 1}^N f\left( {{P_i}} \right){\rm{length}}\left( {{C_i}} \right) = \mathop \sum \limits_{i = 1}^N f\left( {{P_i}} \right)\Delta {s_i}$
where ${P_i}$ is the sample points; and the corresponding $f$ values at these points are listed in the table.
a. the path ${C_1}$ from $A$ to $B$
We consider three regular partitions from $A$ to $B$ such that $\Delta {s_i} = \frac{1}{3}$, for $1 \le i \le 3$.
So,
$\mathop \smallint \limits_{{C_1}}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq f\left( {1,\frac{1}{6},0} \right)\cdot\frac{1}{3} + f\left( {1,\frac{1}{2},0} \right)\cdot\frac{1}{3} + f\left( {1,\frac{5}{6},0} \right)\cdot\frac{1}{3}$
$\mathop \smallint \limits_{{C_1}}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq 1 + 3.3\cdot\frac{1}{3} + 3.6\cdot\frac{1}{3} = 3.3$
b. the path ${C_2}$ from $B$ to $C$
We consider three regular partitions from $B$ to $C$ such that $\Delta {s_i} = \frac{1}{3}$, for $1 \le i \le 3$.
So,
$\mathop \smallint \limits_{{C_2}}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq f\left( {1,1,\frac{1}{6}} \right)\cdot\frac{1}{2} + f\left( {1,1,\frac{1}{2}} \right)\cdot\frac{1}{2} + f\left( {1,1,\frac{5}{6}} \right)\cdot\frac{1}{2}$
$\mathop \smallint \limits_{{C_2}}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq 4.2\cdot\frac{1}{3} + 4.5\cdot\frac{1}{3} + 4.2\cdot\frac{1}{3} = 4.3$
Thus, the line integrals of $f$ along $ABC$ is
$\mathop \smallint \limits_{ABC}^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_{{C_1}}^{} f\left( {x,y,z} \right){\rm{d}}s + \mathop \smallint \limits_{{C_2}}^{} f\left( {x,y,z} \right){\rm{d}}s \simeq 3.3 + 4.3 = 7.6$
2. compute the line integral of ${\bf{F}}$ along $ABC$
The line integral of ${\bf{F}}$ along $ABC$ is estimated by the sum:
$\mathop \sum \limits_{i = 1}^N {\bf{F}}\left( {{P_i}} \right)\cdot\Delta {{\bf{r}}_i}$
where ${P_i}$ is the sample points; and the corresponding ${\bf{F}}$ values for these points are listed in the table.
a. the path ${C_1}$ from $A$ to $B$
We consider three regular partitions from $A$ to $B$ such that $\Delta {s_i} = \frac{1}{3}$, for $1 \le i \le 3$.
Since the path is a line segment along the $y$-axis, so $\Delta {{\bf{r}}_i} = \frac{1}{3}{\bf{j}}$. Thus, $\Delta {{\bf{r}}_i} = \left( {0,\frac{1}{3},0} \right)$.
So,
$\mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \mathop \sum \limits_{i = 1}^3 {\bf{F}}\left( {{P_i}} \right)\cdot\Delta {{\bf{r}}_i}$
$ = {\bf{F}}\left( {1,\frac{1}{6},0} \right)\cdot\left( {0,\frac{1}{3},0} \right) + {\bf{F}}\left( {1,\frac{1}{2},0} \right)\cdot\left( {0,\frac{1}{3},0} \right) + {\bf{F}}\left( {1,\frac{5}{6},0} \right)\cdot\left( {0,\frac{1}{3},0} \right)$
$ = \left( {1,0,2} \right)\cdot\left( {0,\frac{1}{3},0} \right) + \left( {1,1,3} \right)\cdot\left( {0,\frac{1}{3},0} \right) + \left( {2,1,5} \right)\cdot\left( {0,\frac{1}{3},0} \right)$
$ = 0 + \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
b. the path ${C_2}$ from $B$ to $C$
We consider three regular partitions from $B$ to $C$ such that $\Delta {s_i} = \frac{1}{3}$, for $1 \le i \le 3$.
Since the path is a line segment along the $z$-axis, so $\Delta {{\bf{r}}_i} = \frac{1}{3}{\bf{k}}$. Thus, $\Delta {{\bf{r}}_i} = \left( {0,0,\frac{1}{3}} \right)$.
So,
$\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \mathop \sum \limits_{i = 1}^3 {\bf{F}}\left( {{P_i}} \right)\cdot\Delta {{\bf{r}}_i}$
$ = {\bf{F}}\left( {1,1,\frac{1}{6}} \right)\cdot\left( {0,0,\frac{1}{3}} \right) + {\bf{F}}\left( {1,1,\frac{1}{2}} \right)\cdot\left( {0,0,\frac{1}{3}} \right) + {\bf{F}}\left( {1,1,\frac{5}{6}} \right)\cdot\left( {0,0,\frac{1}{3}} \right)$
$ = \left( {3,2,4} \right)\cdot\left( {0,0,\frac{1}{3}} \right) + \left( {3,3,3} \right)\cdot\left( {0,0,\frac{1}{3}} \right) + \left( {5,3,3} \right)\cdot\left( {0,0,\frac{1}{3}} \right)$
$ = \frac{4}{3} + 1 + 1 = \frac{{10}}{3}$
Thus, the line integral of ${\bf{F}}$ along $ABC$ is
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \frac{2}{3} + \frac{{10}}{3} = 4$
