Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{4}{\pi ^4}$
Work Step by Step
We have ${\bf{F}}\left( {x,y,z} \right) = \left( {xy,2,{z^3}} \right)$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$ for $0 \le t \le \pi $.
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\cos t\sin t,2,{t^3}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - \sin t,\cos t,1} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right) = \left( {\cos t\sin t,2,{t^3}} \right)\cdot\left( { - \sin t,\cos t,1} \right)$
$ = - \cos t{\sin ^2}t + 2\cos t + {t^3}$
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^\pi \left( { - \cos t{{\sin }^2}t + 2\cos t + {t^3}} \right){\rm{d}}t$
$ = - \mathop \smallint \limits_0^\pi \cos t{\sin ^2}t{\rm{d}}t + 2\mathop \smallint \limits_0^\pi \cos t{\rm{d}}t + \mathop \smallint \limits_0^\pi {t^3}{\rm{d}}t$
$ = - \mathop \smallint \limits_0^\pi {\sin ^2}t{\rm{d}}\left( {\sin t} \right) + 2\mathop \smallint \limits_0^\pi \cos t{\rm{d}}t + \mathop \smallint \limits_0^\pi {t^3}{\rm{d}}t$
$ = - \frac{1}{3}{\sin ^3}t|_0^\pi + 2\sin t|_0^\pi + \frac{1}{4}{t^4}|_0^\pi = \frac{1}{4}{\pi ^4}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{1}{4}{\pi ^4}$.
