How can I calculate $\iint_S\frac{\bf x}{|{\bf x}|^3}\cdot d{\bf S}$ with a semisphere $S$ not centered at the origin? 
Let $$
F(x,y,z)=\frac{x{\bf i}+y{\bf j}+z{\bf k}}{(x^2+y^2+z^2)^{3/2}}. 
$$
  How can I calculate 
  $$
\iint_SF\cdot d{\bf S}
$$
  where $S$ a the "upper semi-unit-sphere" and  the boundary of $S$ given by
  $$
\begin{cases}
x+y+z=3\\
(x-1)^2+(y-1)^2+(z-1)^2=1
\end{cases}?
$$




*

*If I change the coordinate to make the equations of the boundary of the semisphere simpler:
$$
z'=0,\quad x'^2+y'^2+z'^2=1,
$$
then I messed up with $F(x',y',z')$. But if I don't change the coordinates, I messed up with the parameterization of the surface. Any idea?

*Does the integral $$\iint_S\frac{\bf x}{|{\bf x}|^3}\cdot d{\bf S}$$ have some meaning in physics?



[Added] I didn't expect that my description of the surface in the integral is so difficult to be understood. Suppose we have a unit sphere centered at $(1,1,1)$ 
$$
\Omega=\{(x,y,z)\in{\Bbb R}^3:(x-1)^2+(y-1)^2+(z-1)^2=1\}
$$
and the plane 
$$
P=\{(x,y,z)\in{\Bbb R}^3:x+y+z=3\}.
$$
Geometrically, the plane $P$ would cut the sphere $\Omega$ into two pieces and $S$ is one of them while I don't specify which one so that the result would be up to the choice of these two pieces. 
The confusion might due to my notation:
$$
\begin{cases}
x+y+z=3\\
(x-1)^2+(y-1)^2+(z-1)^2=1
\end{cases}.
$$
which is equivalent to 
$$
\{(x,y,z)\in{\Bbb R}^3:x+y+z=3\ {\bf and}\ (x-1)^2+(y-1)^2+(z-1)^2=1\}
$$
which is the boundary of $S$.

There are a few ways to solve this problem. 


*

*You are asked to evaluate
$$
\int \mathbf{F} \cdot \mathbf{dS}
$$
The normal to the plane is $(1,1,1)3^{-1/2}$ , and hence the integral reduces to 
$$
\int_{\{(x-1)^2+(y-1)^2+(z-1)^2 \leq 1 \cap x+y+z=3\}} \frac{\sqrt{3}}{(x^2 + y^2 + z^2)^{3/2}}\,dS
$$
Now the integrand is rotationally symmetric, and hence one can rotate the region and integrate over a suitable disc with center on the $z-$ axis at $(0,0,\sqrt3).$ 
Specifically the integral is (in polar coordinates)
$$
\int_0^1 \int_0^{2\pi} \frac{\sqrt{3}r\,dr\,d\theta}{(r^2 + 3)^{3/2}}
$$
Note this tells you in particular that the integral is not 0, since the integrand is positive.

*A second method is as follows. You need to evaluate the following surface integral (stokestheorem) over an appropriate portion of a sphere of radius 2, centered at the origin. I'll leave the details to you, one can evaluate this integral by switching to spherical coordinates. 

*A third (essentially same as the second ) approach would be to look at the integral over unit disc on the plane $x+y+z=3$ centered $(1,1,1)$ as equal to (again  by stoke's theorem) the integral over a cone below the said disc, above the origin, with a ball of radius $\epsilon$ around the origin punctured. 
The integral on the surface of the cone vanishes (why?), and you're left to evaluate the integral in (2) except on a sphere centered origin radius $\epsilon.$ The answer will turn out to be indepenedent of $\epsilon.$ 
