Answer
$\mathop \smallint \limits_C^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}s = \sqrt 2 \pi \left( {1 + \frac{1}{3}{\pi ^2}} \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2}$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,t} \right)$ for $0 \le t \le \pi $.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = {\cos ^2}t + {\sin ^2}t + {t^2} = 1 + {t^2}$
$||{\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( {1 + {t^2}} \right){\rm{d}}t$
$ = \sqrt 2 \left( {t + \frac{1}{3}{t^3}} \right)|_0^\pi = \sqrt 2 \left( {\pi + \frac{1}{3}{\pi ^3}} \right)$
So, $\mathop \smallint \limits_C^{} \left( {{x^2} + {y^2} + {z^2}} \right){\rm{d}}s = \sqrt 2 \pi \left( {1 + \frac{1}{3}{\pi ^2}} \right)$.
