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