Answer
$\mathop \smallint \limits_C^{} \left( {xy + z} \right){\rm{d}}s = \frac{1}{2}\sqrt 2 {\pi ^2}$
Work Step by Step
We have $f\left( {x,y,z} \right) = xy + z$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$ for $0 \le t \le \pi $.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = \left( {\cos t} \right)\left( {\sin t} \right) + t = \frac{1}{2}\sin 2t + t$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( { - \sin t,\cos t,1} \right)\cdot\left( { - \sin t,\cos t,1} \right)} $
$ = \sqrt {{{\sin }^2}t + {{\cos }^2}t + 1} = \sqrt 2 $
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$ = \sqrt 2 \mathop \smallint \limits_0^\pi \left( {\frac{1}{2}\sin 2t + t} \right){\rm{d}}t$
$ = \sqrt 2 \left( { - \frac{1}{4}\cos 2t + \frac{1}{2}{t^2}} \right)|_0^\pi $
$ = \sqrt 2 \left( { - \frac{1}{4} + \frac{1}{2}{\pi ^2} + \frac{1}{4}} \right) = \frac{1}{2}\sqrt 2 {\pi ^2}$
So, $\mathop \smallint \limits_C^{} \left( {xy + z} \right){\rm{d}}s = \frac{1}{2}\sqrt 2 {\pi ^2}$.
