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