line integral: anticlockwise parametrisation in $\mathbb R^3$

Question

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?

Answer 1

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 $n={1\over\sqrt{3}}(-1,-1,-1)$.

Answer 2

Usually, "counterclockwise" without further qualification means "counterclockwise from above", i.e. from the positive $z$-direction. However, this should really be made explicit in the setup in order to avoid this type of confusion, especially since "counterclockwise from above" doesn't make sense for all types of curves in $\mathbb R^3$. You should insist that whoever gave you this question be clearer.