You are correct that $F$ is conservative with potential $f(x,y) = xy$, which in polar coordinates can also be written as $\tilde{f}(\theta,r)=r^2\sin\theta\cos\theta$. So, then $\int_C F\cdot dr$ does only depend on the start and end points. I suppose if you wanted to be very clear, you could show that the function you give is the potential of $F$, which is easier to do in Cartesian coordinates.
I hope you were using separate unit normals for each segment!
For segment $(-1,0)$ to $(0,1)$ anticlockwise with outer unit normal $\dfrac1{\sqrt{2}}\langle-1,1\rangle$, $x=-t,y=1-t,t:0\rightarrow1, ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}~dt=\sqrt{2}~dt$,
$\displaystyle\int_C\mathbf{F}\cdot\mathbf{n}=\int_0^1\dfrac1{\sqrt2}(-x+y^2)~ds=\dfrac{\sqrt2}{\sqrt2}\int_0^1t+(1-t)^2~dt=\dfrac56$.
If you keep the same direction and change to the inner unit normal $\dfrac1{\sqrt{2}}\langle1,-1\rangle$, you get $\displaystyle\int_0^1\dfrac1{\sqrt2}(x-y^2)~ds=\int_0^1-t-(1-t)^2~dt=\dfrac{-5}{6}$, i.e. the sign changes.
If you switch direction to clockwise but keep the outer unit normal $\dfrac1{\sqrt{2}}\langle-1,1\rangle$, you get
$x=t-1,y=t,t:0\rightarrow1,ds=\sqrt{2}~dt$, $\displaystyle\int_0^1\dfrac1{\sqrt2}(-x+y^2)~ds=\int_0^11-t+t^2~dt=\dfrac5{6}$
For this kind of "flux across curve" line integral, $\displaystyle\int_C\mathbf{F}\cdot\mathbf{n}~ds$, it does matter which unit normal you use - switching gives you a sign change. It doesn't matter about the direction of the curve. Why not? We're summing $\mathbf{F}\cdot\mathbf{n}$ along the curve, and this quantity does not depend on the direction of traversal (but does depend on the choice of normal). Further, $ds$, the 'piece of curve', is positive regardless of the direction of traversal. For the same reasons, the general line integral $\displaystyle\int_C\phi(x,y,z)~ds$ doesn't depend on the direction of traversal.
You can contrast this with the "work along curve" line integral,
$\displaystyle\int_C\mathbf{F}\cdot d\mathbf{r}$ where $d\mathbf{r}$ does depend on the direction of traversal. It's $d\mathbf{r}$, the instantaneous vector in the direction of traversal, that changes sign with a change in direction of traversal.
In more physical terms:
- the flux integral is calculating the amount of flow perpendicular to the curve. Adding this up doesn't depend on which way you traverse the curve, but does depend on your idea of inside/outside (the choice of normal).
- the work integral is calculating the amount of work needed to get from A to B. It matters on a given part of the curve whether you're moving with the current/wind or against it, and this depends on the direction of travel.
Best Answer
Consider the quarter arc: $\gamma:[0,\pi/2]\to \mathbf{R}^2$ with $$ \gamma(t) = (\cos(t),\sin(t)) $$ with the direction from $\gamma(0)$ to $\gamma(1)$. Then $\int_L Pdx+Qdy=\int_\gamma Pdx+Qdy$ by Green's theorem, where $$ P(x,y)=\frac{-y}{x^2+y^2},\quad Q(x,y)=\frac{x}{x^2+y^2}\;, $$ because the closed path $L-\gamma$ is contained in an open simply connected subset of $\mathbf{R}^2$ where the vector field is smooth and $P_y=Q_x$. By working on the line integral along $\gamma$, you can easily find the expected answer $\pi/2$ mentioned in your post: $$ \int_\gamma Pdx+Qdy=\int_{0}^{\pi/2}\bigg((-\sin(t))(-\sin(t))+\cos(t)\cos(t)\bigg)\;dt = \frac{\pi}{2}\;. $$
The vector field $(P,Q)$ does not have a potential function on the punctured plane $\mathbf{R}^2\setminus\{(0,0)\}$. So when you work on a potential function, you need to specify the (simply connected) domain of your vector field.