Answer
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \frac{{11}}{6}$
Work Step by Step
We have $f\left( {x,y,z} \right) = {{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {\ln t,\sqrt 2 t,\frac{1}{2}{t^2}} \right)$ for $1 \le t \le 2$, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = {{\rm{e}}^{\ln t}} - \frac{{\sqrt 2 t}}{{2\sqrt 2 \left( {\frac{1}{2}{t^2}} \right)}} = t - \frac{1}{t}$
$ds = ||{\bf{r}}'\left( t \right)||dt = \sqrt {\left( {\frac{1}{t},\sqrt 2 ,t} \right)\cdot\left( {\frac{1}{t},\sqrt 2 ,t} \right)} dt$
$ds = \sqrt {\frac{1}{{{t^2}}} + 2 + {t^2}} dt = \frac{1}{t}\sqrt {{t^4} + 2{t^2} + 1} dt$
By Eq. (4) in Section 17.2:
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \mathop \smallint \limits_1^2 \left( {t - \frac{1}{t}} \right)\left( {\frac{1}{t}\sqrt {{t^4} + 2{t^2} + 1} } \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \mathop \smallint \limits_1^2 \left( {\sqrt {{t^4} + 2{t^2} + 1} - \frac{1}{{{t^2}}}\sqrt {{t^4} + 2{t^2} + 1} } \right){\rm{d}}t$
But ${t^4} + 2{t^2} + 1 = {\left( {{t^2} + 1} \right)^2}$. So, $\sqrt {{t^4} + 2{t^2} + 1} = {t^2} + 1$.
The integral becomes
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \mathop \smallint \limits_1^2 \left( {{t^2} + 1} \right){\rm{d}}t - \mathop \smallint \limits_1^2 \frac{1}{{{t^2}}}\left( {{t^2} + 1} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \mathop \smallint \limits_1^2 \left( {{t^2} + 1} \right){\rm{d}}t - \mathop \smallint \limits_1^2 \left( {1 + \frac{1}{{{t^2}}}} \right){\rm{d}}t$
$ = \left( {\frac{1}{3}{t^3} + t} \right)|_1^2 - \left( {t - \frac{1}{t}} \right)|_1^2 = \left( {\frac{8}{3} + 2 - \frac{1}{3} - 1} \right) - \left( {2 - \frac{1}{2} - 1 + 1} \right) = \frac{{11}}{6}$
So, $\mathop \smallint \limits_C^{} \left( {{{\rm{e}}^x} - \frac{y}{{2\sqrt 2 z}}} \right){\rm{d}}s = \frac{{11}}{6}$.
