Answer
We show that the integral of ${\bf{F}}$ along the segment from $P = \left( {a,b} \right)$ to $Q = \left( {a,c} \right)$ is equal to the angle $\angle POQ$.
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y} \right) = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$.
The line segment $\overrightarrow {PQ} $ can be parametrized by
${\bf{r}}\left( t \right) = \left( {a,b} \right) + t\left( {\left( {a,c} \right) - \left( {a,b} \right)} \right)$
$ = \left( {a,b} \right) + t\left( {0,c - b} \right)$
$ = \left( {a,t\left( {c - b} \right) + b} \right)$
for $0 \le t \le 1$.
So,
${\bf{r}}'\left( t \right) = \left( {0,c - b} \right)$
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\frac{{ - \left( {t\left( {c - b} \right) + b} \right)}}{{{a^2} + {{\left( {t\left( {c - b} \right) + b} \right)}^2}}},\frac{a}{{{a^2} + {{\left( {t\left( {c - b} \right) + b} \right)}^2}}}} \right)$
Evaluate the line integral:
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 \left( {\frac{{ - \left( {t\left( {c - b} \right) + b} \right)}}{{{a^2} + {{\left( {t\left( {c - b} \right) + b} \right)}^2}}},\frac{a}{{{a^2} + {{\left( {t\left( {c - b} \right) + b} \right)}^2}}}} \right)\cdot\left( {0,c - b} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \frac{{a\left( {c - b} \right)}}{{{a^2} + {{\left( {t\left( {c - b} \right) + b} \right)}^2}}}{\rm{d}}t$
Write $u = t\left( {c - b} \right) + b$. So, $du = \left( {c - b} \right)dt$.
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = a\mathop \smallint \limits_b^c \frac{1}{{{a^2} + {u^2}}}{\rm{d}}u$
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,
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = {\tan ^{ - 1}}\frac{u}{a}|_b^c = {\tan ^{ - 1}}\frac{c}{a} - {\tan ^{ - 1}}\frac{b}{a}$
But ${\tan ^{ - 1}}\frac{c}{a}$ is the angle between $\overrightarrow {OQ} $ and the $x$-axis; and ${\tan ^{ - 1}}\frac{b}{a}$ is the angle between $\overrightarrow {OP} $ and the $x$-axis. So, ${\tan ^{ - 1}}\frac{c}{a} - {\tan ^{ - 1}}\frac{b}{a} = \theta $, which is the angle $\angle POQ$.
Hence, the integral of ${\bf{F}}$ along the segment from $P = \left( {a,b} \right)$ to $Q = \left( {a,c} \right)$ is equal to the angle $\angle POQ$.
