Line integral of a vector field $$\int_{\gamma}ydx+zdy+xdz$$ 
given that $\gamma$ is the intersection of $x+y=2$ and $x^2+y^2+z^2=2(x+y)$ and its projection in the $xz$ plane is taken clockwise.
In my solution, I solved the non-linear system of equations and I found that $x^2+y^2+z^2 = 4$. Given the projection in the $xz$ plane is taken clockwise, the parametrization is: $\gamma(t) = (2\sin t, 2 - 2\sin t,2\cos t)$, $0<t<2\pi$. But when I evaluate this integral, I keep getting wrong results:
$\int_{0}^{2\pi}[(2-2\cos t)2\cos t+2\cos t(-2\cos t)+2\sin t(-2\sin t)]dt$, I've only susbstituted $x,y,z$ and $dx,dy,dz$; which leads me to $-8\pi$, but the answer on my textbook is $-2\pi \sqrt{2}$.
Is there any conceptual mistake in my solution?

Alternatively, you can use the Stokes theorem here, it simplifies calculations. Let $\vec{F}=(y,z,x)$. Since $\gamma$ is a closed curve, the following equality holds:
$$
\oint_{\gamma}\vec{F}\cdot d\vec{r} = \iint_{S}\nabla\times\vec{F}\cdot d\vec{S}
= \iint_{S}\nabla\times\vec{F}\cdot\vec{n}\;dS,
$$
and this is not a bad idea because $\nabla\times\vec{F}$ simplifies to $-(1,1,1)$ and $\vec{n}$ is easy to compute as the surface $S$ is part of the plane $x+y=2$, in other words $\vec{n}=\frac{1}{\sqrt{2}}(1,1,0)$ (note that it is correctly oriented). Therefore:
$$
\iint_{S}\nabla\times\vec{F}\cdot\vec{n}\;dS = -\sqrt{2}\;Area(S)=-\sqrt{2}\pi2,
$$
 since $S$ is a circle with radius $\sqrt{2}$ (indeed, the center of the sphere ,$(1,1,0)$, lies in the plane $x+y=2$, so the intersection of the plane and the sphere is a circle).
