Computing line integral using Stokes´theorem Use Stokes´ theorem to show that $$\int_C ydx+zdy+xdz=\pi a^2\sqrt{3}$$ where $C$ is the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the plane $x+y+z=0$
My attempt: 
By Stokes´ theorem I know that $$\int_S (\nabla \times F) \cdot n \ dS=\int_c F \cdot  d\alpha$$  In this case the intersection curve $C$ is a circle and $S$ is "half" of the sphere 
using $r(u,v)=(acos(u)sin(v), a sin(u)cos(u), acos(v))$ $0\le v \le \pi$, $-\pi/4 \le u \le 3\pi/4$ as a parametrization of the sphere and computing $\nabla \times F=(-1,-1,-1)$ (where $F=(y,z,x)$), and $${\partial r\over \partial u}\times {\partial r\over \partial v}$$ the surface integral becomes:
$$\int_{-\pi/4}^{3\pi/4}\int_{0}^{\pi}({a^2sin(v)sin(2u)\over 2}-a^2sin(v)sin^2(u)+{a^2sin(2v)\over 2})dv\ du$$, but after computing the integral I don´t get the answer, Can you please tell me where is my mistake?

You did not specify the orientation of the curve $C$. But, you gave the answer. So we have to choose the orientation that will give the given solution.  The curve $C$ is in the plane $x+y+z = 0$.  The vector $(-1,-1,-1)$ determines one orientation of this plane.  Let the curve $C$ be oriented using the right-hand rule.  Let $D$ be the disk in the plane $x+y+z = 0$  bounded by the circle $C$; it is the disk cetered at the origin and with radius $a$.  The unit normal to $D$ is $(-1,-1,-1)/\sqrt{3}$. By Stokes' theorem
\begin{align*}
\int_C (y dx+z dy + x dz) & = \iint_D (-1,-1,-1)\cdot (-1,-1,-1)/\sqrt{3} \ dA \\
 & = \sqrt{3}  \iint_D dA \\
 & = \pi a^2 \sqrt{3},
\end{align*}
since $\iint_D dA$ is the area of the disk $D$. 
