Answer
$\mathop \smallint \limits_C^{} \left( {x + 2y + z} \right){\rm{d}}s = 3\sqrt 2 + \frac{{\sqrt 2 }}{8}{\pi ^2}$
Work Step by Step
We have $f\left( {x,y,z} \right) = x + 2y + z$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$ for $0 \le t \le \frac{\pi }{2}$, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = \cos t + 2\sin t + t$
$ds = ||{\bf{r}}'\left( t \right)||dt = \sqrt {\left( { - \sin t,\cos t,1} \right)\cdot\left( { - \sin t,\cos t,1} \right)} dt$
$ds = \sqrt {{{\sin }^2}t + {{\cos }^2}t + 1} dt = \sqrt 2 dt$
By Eq. (4) in Section 17.2:
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {x + 2y + z} \right){\rm{d}}s = \sqrt 2 \mathop \smallint \limits_0^{\pi /2} \left( {\cos t + 2\sin t + t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {x + 2y + z} \right){\rm{d}}s = \sqrt 2 \left( {\sin t - 2\cos t + \frac{1}{2}{t^2}} \right)|_0^{\pi /2}$
$ = \sqrt 2 \left( {1 + \frac{1}{8}{\pi ^2} + 2} \right) = 3\sqrt 2 + \frac{{\sqrt 2 }}{8}{\pi ^2}$
So, $\mathop \smallint \limits_C^{} \left( {x + 2y + z} \right){\rm{d}}s = 3\sqrt 2 + \frac{{\sqrt 2 }}{8}{\pi ^2}$.