Answer
$\mathop \smallint \limits_C^{} \sqrt {1 + 9xy} {\rm{d}}s = \frac{{14}}{5}$
Work Step by Step
We have $f\left( {x,y} \right) = \sqrt {1 + 9xy} $. The curve $y = {x^3}$ for $0 \le x \le 1$ can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,{t^3}} \right)$ for $0 \le t \le 1$.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = \sqrt {1 + 9{t^4}} $
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,3{t^2}} \right)\cdot\left( {1,3{t^2}} \right)} = \sqrt {1 + 9{t^4}} $
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_0^1 \left( {1 + 9{t^4}} \right){\rm{d}}t = \left( {t + \frac{9}{5}{t^5}} \right)|_0^1 = 1 + \frac{9}{5} = \frac{{14}}{5}$
So, $\mathop \smallint \limits_C^{} \sqrt {1 + 9xy} {\rm{d}}s = \frac{{14}}{5}$.