Answer
(a) $f\left( {{\bf{r}}\left( t \right)} \right) = 3{t^2} + {t^4}$
$ds = 2\sqrt {11} tdt$
(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 ${\bf{r}}\left( t \right) = \left( {3{t^2},{t^2},{t^2}} \right)$ for $0 \le t \le \sqrt 2 $, we obtain
$f\left( {{\bf{r}}\left( t \right)} \right) = 3{t^2} + {t^4}$
$ds = ||{\bf{r}}'\left( t \right)||dt$
$ds = \sqrt {\left( {6t,2t,2t} \right)\cdot\left( {6t,2t,2t} \right)} dt = \sqrt {44} tdt = 2\sqrt {11} tdt$
(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^{\sqrt 2 } \left( {3{t^2} + {t^4}} \right)t{\rm{d}}t$
$ = 2\sqrt {11} \mathop \smallint \limits_0^{\sqrt 2 } \left( {3{t^3} + {t^5}} \right){\rm{d}}t$
$ = 2\sqrt {11} \left( {\frac{3}{4}{t^4} + \frac{1}{6}{t^6}} \right)|_0^{\sqrt 2 }$
$ = 2\sqrt {11} \left( {3 + \frac{8}{6}} \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} $.
Notice that it gives the same result as in Exercise 1.
