<s>[INST] a surface integral problem $F=(y-z,z-x,x-y)$ $S$ be the portion of surface defined by  $x^2+y^2+z^2=1$ and $x+y+z\ge 1$. We want to evaluate $\int_S curl(F)\cdot dS$.
I have found $curl(F)=(-2,-2,-2)$ and the normal vector of $S$ is $(x,y,z)$. Thus the integral becomes $\int_S -2(x+y+z) dS$. However, if I use the polar coordinate of $S$, then the boundary of $\phi$ and $\theta$ is a big problem. 
I still consider using Stoke's Theorem.  It then should be $\int_A F\cdot dr$ where $A$ is the circle defined by the unit ball and the plane $x+y+z=1$. However I can not find a way to paramettrize this circle. That is a big problem.
 [/INST] You can also use the divergence theorem here, it will save you the heavy computations you are not comfortable with:
If $S_2$ denotes the part of the plane $x+y+z=1$ inside the sphere $x^2+y^2+z^2=1$, and $E$ the 3D-region bounded by $S$ and $S_2$, then
$$
\iint_S \nabla \times F\cdot dS+ \iint_{S_2} \nabla \times F\cdot dS= \iiint_E \nabla\cdot \nabla \times F\; dV = 0
$$
Since the plane $x+y+z=1$ has normal unitary vector $n=-\frac{\pmatrix{1\\1\\1} }{\sqrt{3}}$, it follows that
$$
\iint_S \nabla \times F\cdot dS= - \iint_{S_2} \nabla \times F\cdot dS = - \iint_{S_2}\pmatrix{-2\\-2\\-2}\cdot\pmatrix{-1\\-1\\-1}\frac{dS}{\sqrt{3}}=\frac{-6}{\sqrt{3}}\;A(S_2)=\frac{-6}{\sqrt{3}}\; \frac{2\pi}{3}=\frac{-4 \pi}{\sqrt{3}}
$$

Note: Perhaps the most difficult part is finding the area of $S_2$ ($A(S_2)$). The projection of $S_2$ in the $xy$ plane is the ellipse $x^2+y^2+(1-x-y)^2=1$, which has area $A'=\frac{2\pi}{3\sqrt{3}}$. It is not difficult to see that $A(S_2)=\sqrt{3}A'$. Alternatively, you can use the arguments described  here to show that the disc $S_2$ has radius $\sqrt{\frac{2}{3}}$.
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