Answer
1. For the path ${{\bf{r}}_1} = \left( {t,t,t} \right)$ for $0 \le t \le 1$
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 3$
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{{\bf{r}}_1}\left( 1 \right)} \right) - f\left( {{{\bf{r}}_1}\left( 0 \right)} \right) = 3$
Both methods give the same answer.
2. For the path ${{\bf{r}}_2} = \left( {t,{t^2},{t^3}} \right)$ for $0 \le t \le 1$
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 3$
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{{\bf{r}}_2}\left( 1 \right)} \right) - f\left( {{{\bf{r}}_2}\left( 0 \right)} \right) = 3$
Both methods give the same answer.
Work Step by Step
1. For the path ${{\bf{r}}_1} = \left( {t,t,t} \right)$ for $0 \le t \le 1$
So, we have $d{\bf{r}} = \left( {1,1,1} \right)dt$.
a. determine $\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ directly
From $f = zy + xy + xz$ we get ${\bf{F}} = \nabla f = \left( {y + z,z + x,y + x} \right)$. So,
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{t = 0}^1 \left( {y + z,z + x,y + x} \right)\cdot{\rm{d}}{\bf{r}}$
$ = \mathop \smallint \limits_{t = 0}^1 \left( {2t,2t,2t} \right)\cdot\left( {1,1,1} \right)dt$
$ = \mathop \smallint \limits_{t = 0}^1 6tdt = 3{t^2}|_0^1 = 3$
b. using Theorem 1, we get
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{{\bf{r}}_1}\left( 1 \right)} \right) - f\left( {{{\bf{r}}_1}\left( 0 \right)} \right) = f\left( {1,1,1} \right) - f\left( {0,0,0} \right) = 3$
So both methods give the same answer.
2. For the path ${{\bf{r}}_2} = \left( {t,{t^2},{t^3}} \right)$ for $0 \le t \le 1$
So, we have $d{\bf{r}} = \left( {1,2t,3{t^2}} \right)dt$.
a. determine $\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ directly
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{t = 0}^1 \left( {y + z,z + x,y + x} \right)\cdot{\rm{d}}{\bf{r}}$
$ = \mathop \smallint \limits_{t = 0}^1 \left( {{t^2} + {t^3},t + {t^3},t + {t^2}} \right)\cdot\left( {1,2t,3{t^2}} \right)dt$
$ = \mathop \smallint \limits_{t = 0}^1 \left( {{t^2} + {t^3} + 2{t^2} + 2{t^4} + 3{t^3} + 3{t^4}} \right)dt$
$ = \mathop \smallint \limits_{t = 0}^1 \left( {5{t^4} + 4{t^3} + 3{t^2}} \right)dt = \left( {{t^5} + {t^4} + {t^3}} \right)|_0^1 = 3$
b. using Theorem 1, we get
$\mathop \smallint \limits_C {\bf{F}}\cdot{\rm{d}}{\bf{r}} = f\left( {{{\bf{r}}_2}\left( 1 \right)} \right) - f\left( {{{\bf{r}}_2}\left( 0 \right)} \right) = f\left( {1,1,1} \right) - f\left( {0,0,0} \right) = 3$
So both methods give the same answer.
