Answer
(a) $f\left( {{\bf{r}}\left( t \right)} \right) = 6t + 4{t^2}$
$ds = 2\sqrt {11} dt$
(b) $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \frac{{26}}{3}\sqrt {11} $
Work Step by Step
(a) We have $f\left( {x,y,z} \right) = x + yz$.
Using the parametrization ${\bf{r}}\left( t \right) = \left( {6t,2t,2t} \right)$ for $0 \le t \le 1$, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = 6t + 4{t^2}$
$ds = ||{\bf{r}}'\left( t \right)||dt = \sqrt {\left( {6,2,2} \right)\cdot\left( {6,2,2} \right)} dt = \sqrt {44} dt = 2\sqrt {11} dt$
(b) By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$ = 2\sqrt {11} \mathop \smallint \limits_0^1 \left( {6t + 4{t^2}} \right){\rm{d}}t$
$ = 2\sqrt {11} \left( {3{t^2} + \frac{4}{3}{t^3}} \right)|_0^1$
$ = 2\sqrt {11} \left( {3 + \frac{4}{3}} \right) = \frac{{26}}{3}\sqrt {11} $
So, $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \frac{{26}}{3}\sqrt {11} $.
