Answer
The flux across the curve is $18$.
Work Step by Step
The line segment $C$ from $\left( {3,0} \right)$ to $\left( {0,3} \right)$, oriented upward, can be parametrized by
${\bf{r}}\left( t \right) = \left( {3,0} \right) + t\left( {\left( {0,3} \right) - \left( {3,0} \right)} \right)$
${\bf{r}}\left( t \right) = \left( { - 3t + 3,3t} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{{\left( { - 3t + 3} \right)}^2},9{t^2}} \right)$
The vectors that are normal to $C$ is given by
${\bf{N}}\left( t \right) = \left( {y'\left( t \right), - x'\left( t \right)} \right) = \left( {3,3} \right)$
Using Eq. (10), we calculate the flux across $C$, oriented clockwise:
Flux across $C = $ $\mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{N}}\left( t \right){\rm{d}}t$
Flux across $C = $ $\mathop \smallint \limits_0^1 \left( {{{\left( { - 3t + 3} \right)}^2},9{t^2}} \right)\cdot\left( {3,3} \right){\rm{d}}t$
Flux across $C = $ $\mathop \smallint \limits_0^1 \left( {27{t^2} - 54t + 27 + 27{t^2}} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {54{t^2} - 54t + 27} \right){\rm{d}}t$
$ = 18{t^3} - 27{t^2} + 27t|_0^1 = 18$
So, the flux across $C$ is $18$.
