Sign problem with integration by parts I have a problem with a minus sign in the integration formula, but I don't know where it comes from.
Consider the domain $Q=\{(x,y)\in \mathbb R^2|x^2+y^2<1,y<0\}$, ie the lower half disk. Consider as well a $C^\infty$ function $g$ such that $g\equiv 0$ inside a neighborhood of the circumference $S=\{(x,y)|x^2+y^2=1\}$.
I want to compute the integral $\int_Q g_{y}$. With the usual orientation of $\mathbb R^2$ and using Fubini we have:
$\int_Q g_{y}=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^0g_{y}(x,y)dydx=\int_{-1}^1g(x,0)dx$
However we can also compute this integral integrating by parts and we get:
 $\int_Q g_{y}=-\int_Q 0\cdot g+\int_{\partial Q} 1 g\nu_2$
where $\nu_2$ is an outer normal to the boundary. On the boundary, $g$ is zero outside the axis so we only integrate on it, there the outer normal is $\nu=(0,1)$ so $\nu_2=1$ and the orientation of the border is from right to left since it has the induced orientation. So we get:
$\int_Q g_{y}=\int_{1}^{-1}g(x,0)dx$
Which has clearly the opposite sign. Where is the problem?

The mistake you are making is in your your assumption that
$$
\int_{\partial Q} g=\int_{1}^{-1}g(x,0)dx.
$$
That formula is wrong. Just writing $\int_1^{-1}$ does not make sense. To calculate the line integral, you have to parametrize the segment; the integral on the right should be
$$
\int_{-1}^1g(t,0)dt
$$
The easiest way for me to see this is to notice that the integration by parts formula, in your case, is nothing but Green's Theorem. 
Let $F(x,y)=(-g(x,y),0)$. Then $\nabla\cdot F\cdot\check k=0-(-g_y)=g_y$ (this is the integrand in Green). 
So 
$$
\int_Qg_y=\int_{\partial Q}F\cdot d\vec s
$$
As you said, $F$ is non-zero only on $y=0$. The parametrization of the segment $[-1,1]$ in counterclockwise way is $(-t,0)$ for $t\in[-1,1]$. So 
$$
\int_{Q}g_y=\int_{\partial Q}F\cdot d\vec s=\int_{-1}^1(-g(-t,0),0)\cdot(-1,0)\,dt=\int_{-1}^1g(-t,0)dt=\int_{-1}^1g(x,0)dx
$$
