Answer
The flux across the parabola is ${\rm{e}} - 1$.
Work Step by Step
The parabola $y = {x^2}$ for $0 \le x \le 1$, oriented left to right can be parametrized by
${\bf{r}}\left( t \right) = \left( {t,{t^2}} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{{\rm{e}}^{{t^2}}},2t - 1} \right)$
The vectors that are normal to the parabola is given by
${\bf{N}}\left( t \right) = \left( {y'\left( t \right), - x'\left( t \right)} \right) = \left( {2t, - 1} \right)$
Using Eq. (10), we calculate the flux across the parabola, oriented left to right:
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 = $ $\mathop \smallint \limits_0^1 \left( {{{\rm{e}}^{{t^2}}},2t - 1} \right)\cdot\left( {2t, - 1} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {2t{{\rm{e}}^{{t^2}}} - 2t + 1} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 {{\rm{e}}^{{t^2}}}{\rm{d}}\left( {{t^2}} \right) - 2\mathop \smallint \limits_0^1 t{\rm{d}}t + \mathop \smallint \limits_0^1 {\rm{d}}t$
$ = \left( {{{\rm{e}}^{{t^2}}}|_0^1 - {t^2}|_0^1 + t|_0^1} \right)$
$ = {\rm{e}} - 1 - 1 + 1$
$ = {\rm{e}} - 1$
So, the flux across the parabola is ${\rm{e}} - 1$.
