Answer
The flux across the line segment is $0.54$.
Work Step by Step
The line segment $C$: $1 \le y \le 4$, which is oriented upward, can be parametrized by
${\bf{r}}\left( t \right) = \left( {0,t} \right)$, ${\ \ \ }$ for $1 \le t \le 4$
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{1}{{1 + {t^2}}},\frac{t}{{1 + {t^2}}}} \right)$
The vectors that are normal to $C$, are pointing along the x-axis, so is given by
${\bf{N}}\left( t \right) = \left( {1,0} \right)$
Using Eq. (10), we calculate the flux across $C$, oriented upward:
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 = $ $\mathop \smallint \limits_1^4 \left( {\frac{1}{{1 + {t^2}}},\frac{t}{{1 + {t^2}}}} \right)\cdot\left( {1,0} \right){\rm{d}}t$
$ = \mathop \smallint \limits_1^4 \frac{1}{{1 + {t^2}}}{\rm{d}}t$
From the Table of Integrals at the end of the book, we know that $\smallint \frac{1}{{{a^2} + {u^2}}}{\rm{d}}u = \frac{1}{a}{\tan ^{ - 1}}\frac{u}{a} + C$. So,
Flux across $C = $ ${\tan ^{ - 1}}t|_1^4 = {\tan ^{ - 1}}4 - \frac{\pi }{4} \simeq 0.54$
So, the flux across $C$ is $0.54$.
