Answer
The flux across the curve is $0$.
Work Step by Step
The upper half of the unit circle $C$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$ ${\ \ \ }$ for $0 \le t \le \pi $
So, ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( { - \sin t,\cos t} \right)$
The vectors that are normal to $C$ is given by
${\bf{N}}\left( t \right) = \left( {y'\left( t \right), - x'\left( t \right)} \right) = \left( {\cos t,\sin t} \right)$
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 $C =$ $\mathop \smallint \limits_\pi ^0 \left( { - \sin t,\cos t} \right)\cdot\left( {\cos t,\sin t} \right){\rm{d}}t = 0$
So, the flux across $C$ is $0$.
