Answer
- the line integral of $f$ along $ABC$
$\mathop \smallint \limits_{ABC}^{} f\left( {x,y} \right){\rm{d}}s \approx \frac{\pi }{4}$
- the line integral of ${\bf{F}}$ along $ABC$
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq 1.585$
Work Step by Step
1. compute the line integral of $f$ along $ABC$
The line integral of $f$ along $ABC$ is estimated by the sum:
$\mathop \sum \limits_{i = 1}^N f\left( {{P_i}} \right){\rm{length}}\left( {{C_i}} \right) = \mathop \sum \limits_{i = 1}^N f\left( {{P_i}} \right)\Delta {s_i}$
where ${P_i}$ is the sample points; and the corresponding $f$ values at these points are listed in the table.
From Figure 18 we obtain the radius $r = \frac{1}{2}$ and the element of the arc length $\Delta {s_i} = r\cdot\Delta {\theta _i} = \frac{1}{2}\cdot\frac{\pi }{6} = \frac{\pi }{{12}}$ for $1 \le i \le 3$.
$\mathop \smallint \limits_{ABC}^{} f\left( {x,y} \right){\rm{d}}s \approx \mathop \sum \limits_{i = 1}^N f\left( {{P_i}} \right)\Delta {s_i} = \frac{\pi }{{12}}\left( {f\left( A \right) + f\left( B \right) + f\left( C \right)} \right)$
$\mathop \smallint \limits_{ABC}^{} f\left( {x,y} \right){\rm{d}}s \approx \frac{\pi }{{12}}\left( {1 - 2 + 4} \right) = \frac{\pi }{4}$
2. compute the line integral of ${\bf{F}}$ along $ABC$
The line integral of ${\bf{F}}$ along $ABC$ is estimated by the sum:
$\mathop \sum \limits_{i = 1}^N {\bf{F}}\left( {{P_i}} \right)\cdot\Delta {{\bf{r}}_i}$
where ${P_i}$ is the sample points; and the corresponding ${\bf{F}}$ values for these points are listed in the table.
The path $ABC$ can be parametrized by ${\bf{r}}\left( t \right) = \left( {\frac{1}{2}\cos t,\frac{1}{2}\sin t} \right)$ for $0 \le t \le \frac{\pi }{2}$.
The vector element of distance is approximated by
$\Delta {{\bf{r}}_i} \simeq {\bf{T}}\left( {{P_i}} \right)\Delta {s_i}$,
where ${\bf{T}}\left( {{P_i}} \right)$ is the unit tangent vector at ${P_i}$.
We evaluate the unit tangent vector:
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}}$
${\bf{T}}\left( t \right) = \frac{{\left( { - \frac{1}{2}\sin t,\frac{1}{2}\cos t} \right)}}{{\sqrt {\left( { - \frac{1}{2}\sin t,\frac{1}{2}\cos t} \right)\cdot\left( { - \frac{1}{2}\sin t,\frac{1}{2}\cos t} \right)} }} = \frac{1}{2}\frac{{\left( { - \sin t,\cos t} \right)}}{{\sqrt {\frac{1}{4}\left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right)} }}$
${\bf{T}}\left( t \right) = \left( { - \sin t,\cos t} \right)$
Whereas the length of the arc length is $\Delta {s_i} = r\cdot\Delta {\theta _i} = \frac{1}{2}\cdot\frac{\pi }{6} = \frac{\pi }{{12}}$ for $1 \le i \le 3$. So,
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \mathop \sum \limits_{i = 1}^3 {\bf{F}}\left( {{P_i}} \right)\cdot\Delta {{\bf{r}}_i}$
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \frac{\pi }{{12}}{\bf{F}}\left( A \right)\cdot\left( { - \sin \frac{\pi }{{12}},\cos \frac{\pi }{{12}}} \right) + \frac{\pi }{{12}}{\bf{F}}\left( B \right)\cdot\left( { - \sin \frac{\pi }{4},\cos \frac{\pi }{4}} \right)$
${\ \ \ \ \ \ \ }$ $ + \frac{\pi }{{12}}{\bf{F}}\left( C \right)\cdot\left( { - \sin \frac{{5\pi }}{{12}},\cos \frac{{5\pi }}{{12}}} \right)$
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq \frac{\pi }{{12}}[\left( {1,2} \right)\cdot\left( { - \sin \frac{\pi }{{12}},\cos \frac{\pi }{{12}}} \right) + \left( {1,3} \right)\cdot\left( { - \sin \frac{\pi }{4},\cos \frac{\pi }{4}} \right)$
${\ \ \ \ \ \ \ }$ $ + \left( { - 2,4} \right)\cdot\left( { - \sin \frac{{5\pi }}{{12}},\cos \frac{{5\pi }}{{12}}} \right)]$
$\mathop \smallint \limits_{ABC}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} \simeq 1.585$
