line integral: anticlockwise parametrisation in $\mathbb R^3$ Consider $\gamma$ given by the sides of the triangle with vertices $(0,0,1)^t$, $(0,1,0)^t$ and $(1,0,0)^t$. So $\gamma$ runs through the sides of the triangle. Let $f(x,y,z)=(y,xz,x^2)$.
I want to calculate $\int_\gamma f(\gamma(t))\cdot\dot\gamma(t)\,\mathrm{d}t$.
So I have to calculate this integral for every side of the triangle.
I know I have to go anticlockwise. So do we get $\gamma_1(s)=(1,0,0)^ts+(0,1,0)^t(1-s)$ or $\gamma_1(s)=(0,1,0)^ts+(1,0,0)^t(1-s)$ for $s\in[0,1]$? From where do you consider the "anticlockwise direction"?
I would guess we are looking from $(0,0,0)^t$. But if you're looking from outside the triangle we get the other way around.
I am a bit confussed. In one or two dimensions it's easy to see but in higher dimensions not. So from where are you looking at the triangle?

Three points in ${\mathbb R}^3$ determine a triangle $T$, but not its "positive normal", nor the sense of direction of $\gamma:=\partial T$. Using Occam's razor I'd assume that the intended orientation of $T$, resp., $\gamma$ is such that $\gamma:=\gamma_1+\gamma_2+\gamma_3$ with
$$\eqalign{\gamma_1(s)&=(0,0,1)(1-s)+(0,1,0)s,\cr
\gamma_2(s)&=(0,1,0)(1-s)+(1,0,0)s,\cr
\gamma_3(s)&=(1,0,0)(1-s)+(0,0,1)s.\cr}
$$
The induced orientation of the surface of $T$ would then be such that the positive normal points to the origin, i.e., one has $n={1\over\sqrt{3}}(-1,-1,-1)$.
