Answer
(a) ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^2},3 + 5{t^2},3 - {t^2}} \right)$
$d{\bf{r}} = \left( {10t, - 2t,1} \right)dt$
(b) ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( { - {t^2} - 6t + 3} \right)dt$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{{26}}{3}$
Work Step by Step
(a) We have ${\bf{F}}\left( {x,y,z} \right) = \left( {{z^2},x,y} \right)$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {3 + 5{t^2},3 - {t^2},t} \right)$, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^2},3 + 5{t^2},3 - {t^2}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {10t, - 2t,1} \right)dt$
(b) Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {{t^2},3 + 5{t^2},3 - {t^2}} \right)\cdot\left( {10t, - 2t,1} \right)dt$
$ = \left( {10{t^3} - 6t - 10{t^3} + 3 - {t^2}} \right)dt$
$ = \left( { - {t^2} - 6t + 3} \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_0^2 \left( { - {t^2} - 6t + 3} \right){\rm{d}}t = \left( { - \frac{1}{3}{t^3} - 3{t^2} + 3t} \right)|_0^2$
$ = - \frac{8}{3} - 12 + 6 = - \frac{{26}}{3}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \frac{{26}}{3}$.
