Answer
$\mathop \smallint \limits_C^{} {z^2}{\rm{d}}s = \frac{{128}}{3}\sqrt {29} $
Work Step by Step
We have $f\left( {x,y,z} \right) = {z^2}$ and the parametrization ${\bf{r}}\left( t \right) = \left( {2t,3t,4t} \right)$ for $0 \le t \le 2$.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = 16{t^2}$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {2,3,4} \right)\cdot\left( {2,3,4} \right)} = \sqrt {4 + 9 + 16} = \sqrt {29} $
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^{} {z^2}{\rm{d}}s = 16\sqrt {29} \mathop \smallint \limits_0^2 {t^2}{\rm{d}}t = \frac{{16}}{3}\sqrt {29} {t^3}|_0^2 = \frac{{128}}{3}\sqrt {29} $
So, $\mathop \smallint \limits_C^{} {z^2}{\rm{d}}s = \frac{{128}}{3}\sqrt {29} $.
