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