Answer
(a) ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^{ - 2}},{t^2}} \right)$
$d{\bf{r}} = \left( {1, - {t^{ - 2}}} \right)dt$
(b) ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {{t^{ - 2}} - 1} \right)dt$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{1}{2}$
Work Step by Step
(a) We have ${\bf{F}} = \left( {{y^2},{x^2}} \right)$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {t,{t^{ - 1}}} \right)$ for $1 \le t \le 2$, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^{ - 2}},{t^2}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {1, - {t^{ - 2}}} \right)dt$
(b) Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {{t^{ - 2}},{t^2}} \right)\cdot\left( {1, - {t^{ - 2}}} \right)dt = \left( {{t^{ - 2}} - 1} \right)dt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using 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_1^2 \left( {{t^{ - 2}} - 1} \right){\rm{d}}t = \left( { - {t^{ - 1}} - t} \right)|_1^2 = \left( { - \frac{1}{2} - 2 + 1 + 1} \right) = - \frac{1}{2}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{1}{2}$.
