Flux Through a Closed Curve - Orientation I want to compute 
$$\int_{C}\boldsymbol{F}\cdot\boldsymbol{n}\, ds\qquad\quad \boldsymbol{F}=\langle x, y^2\rangle$$
where $C$ is the curve given by the triangle with vertices $(-1,0)$, $(0,1)$ and $(1,0)$. I broke up the triangle in the three line segments and computed the line integral of the line segment from $(-1,0)$ to $(0,1)$ and got $5/6$. I did the same for the line segment in the first quadrant and got the same. Funny thing is that I got the same results for each line integral even though I first parametrized each segment counterclockwise and then clockwise, and in both cases I used the outward unit normal $\frac{1}{\sqrt{2}}\langle -1,1\rangle$. 
I thought that when the parametrization is counterclockwise (or clockwise), you have to use outer unit normal (inner unit normal). Is this true, or does the orientation of the curve does not matter? Or may be it does not matter in this particular problem? Thanks!
 A: 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.

