Stokes' theorem application The curve $C$ is made from the three circular arcs, centred at the origin having radius $1$, joined end to end, and lying in the planes $z=0$, $y=-x$ and $y=\frac1{\sqrt3}x$ respectively.
The question I need to solve is this:

Using Stokes' theorem (or otherwise), find $$\oint_C (z^2x,x^2y,y^2z)\cdot d{\mathbf r},$$
going in the anticlockwise direction.

I used mathematica to draw the curve $C$,  it is the boundary of this shape: 
So far, I used Stoke's theorem to deduce that $$\oint_C\mathbf F\cdot d\mathbf r = \int_S\operatorname{curl}\boldsymbol F\cdot\mathbf n\,dS$$
where $S$ is the surface from the picture, which changes the integral into
$$2\int_S (yz, xz, xy)\cdot \mathbf n\,dS.$$
Now I'm not sure how to proceed, I was thinking of using Gauss' theorem, but that requires a closed surface. Should I compute $\mathbf n$, or is there a nice integral theorem I can use?
 A: Define $S$ as the trace of the parametric surface $\vec{r}:\big[-\frac{\pi}{4},\frac{\pi}{6}\big]\times\big[0,\frac{\pi}{2}\big]\rightarrow \mathbb{R}$ by $$\vec{r}(\theta,\phi)=\Big(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi)\Big)$$ The image of $\vec{r}$ is precisely the portion of the unit sphere $\rho=1$ on the restricted domain $(\theta,\phi)\in\big[-\frac{\pi}{4},\frac{\pi}{6}\big]\times\big[0,\frac{\pi}{2}\big]$. Because the unit sphere has equation $x^2+y^2+z^2=1$ we see by examining the gradient of $(x,y,z)\rightarrow x^2+y^2+z^2$ that $\vec{n}=\big<x,y,z\big>$ is the outward$-$pointing unit normal of $S$ at the point $(x,y,z)\in S$. Since $\mathrm{d}S=\|\vec{r}_\theta\times \vec{r}_\phi\|\mathrm{d}A=\sin(\phi)\mathrm{d}A$ and $\partial S=C$ we have
$$\begin{eqnarray*}\oint_{C}(xz^2,x^2y,y^2z)\cdot \mathrm{d}\vec{r}&=&\iint_{S}\Big[2\big<yz,xz,xy\big>\cdot\big<x,y,z\big>\Big]\mathrm{d}S \\&=& \iint_{S}6xyz\mathrm{d}S \\&=& \int_0^{\pi/2}\int_{-\pi/4}^{\pi/6}\sin^3(\phi)\cos(\phi)\sin(\theta)\cos(\theta)\mathrm{d}\theta\mathrm{d}\phi \\ &=& -\frac{3}{16}\end{eqnarray*}$$ Note we're using the outward pointing unit normal and not the inward pointing as the outward pointing unit normal induces an anti$-$clockwise orientation on $C$.
