Line Integral of a circle I have this question from "Div,Grad,Curl and all that" by Schey
I have: $$F =e_{\theta}/r$$ 
Find the line integral F t from the point $P_{1}(0,-1,0)$ to point $P_{2}(0,1,0)$ over two difference paths:$C_{R}$, the right hand side of the circle of radius 1 lying on xy plane centered at the origin, and $C_{L}$, the left hand side of the same circle.
The problem is, the question states that the result of line integration over the two different paths ($C_{R}$ and $C_{L}$) is not the same (path dependent line integral); but my calculations shows that it is path independent i.e. Line integration over path $C_{R}$ = line integration over path $C_{L}$...!? Am I wrong?!

 A: Very likely you did something like this:

On the right arc $P_1P_2$, $\mathrm{d}\vec\ell=\vec{e}_\theta\mathrm{d}\theta$, hence
  $$\vec F\cdot\mathrm{d}\vec\ell=1.$$
  On this path, $\theta$ ranges from $-\pi/2$ to $\pi/2$, hence the circulation is $\pi$.
On the left arc $P_1P_2$, $\mathrm{d}\vec\ell=-\vec{e}_\theta\mathrm{d}\theta$, hence
  $$\vec F\cdot\mathrm{d}\vec\ell=-1.$$
  On this path, $\theta$ ranges from $3\pi/2$ to $\pi/2$ (it's decreasing!), hence the circulation is $-1\times(\pi/2-3\pi/2)=-1\times(-\pi)=\pi$.


Let's recall the general method to compute the circulation of a vector field along a path: the first thing you need is a parametrization of this path, say
$$[a,b]\longrightarrow\mathbb{R}^2:t\longmapsto\bigl(x(t),y(t)\bigr).$$
(with $a<b$, and that's important).
Then, given a vector field $\vec F$, the circulation is given by
$$\mathscr{C}=\int_a^b\vec F\bigl(x(t),y(t)\bigr)\cdot\bigl(x'(t),y'(t)\bigr)\,\mathrm{d}t.$$

Now in your case, the left arc $P_1P_2$ is parametrized by
$$\begin{cases}x(t)=\cos(t)\\y(t)=\sin(t)\end{cases}\qquad t\in[-\pi/2,\pi/2].$$
Observe that $\bigl(x'(t),y'(t)\bigr)=\vec e_\theta$
hence
$$\mathscr{C}=\int_{-\pi/2}^{\pi/2}\,\mathrm{d}\theta=\pi$$
(since on this path, $r=1$).
The right arc $P_1P_2$ is parametrized by
$$\begin{cases}x(t)=-\cos(t)\\y(t)=\sin(t)\end{cases}\qquad t\in[-\pi/2,\pi/2].$$
Observe that $\bigl(x'(t),y'(t)\bigr)=-\vec e_\theta$
hence
$$\mathscr{C}=-\int_{-\pi/2}^{\pi/2}\,\mathrm{d}\theta=-\pi$$
(since on this path, $r=1$).

Try to see where (and why) your reasoning was wrong.
A: I think that the integral over the right is 
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d
\theta=\pi$$ and the left is 
$$\int_{-\frac{\pi}{2}}^{-\frac{3\pi}{2}}d
\theta=-\pi$$
