What is the physical meaning of this integral? Let $$I=\int_S ~z~dS$$
Where $S$ is the surface of a hemisphere with equation $x^2+y^2+z^2=4~~~~z \geq0$.
I know $$\int_S~dS$$ would be the surface area of the hemisphere but I can't figure out how the $z$ would change this?
 A: The integral is the avarage of values of the $z$-coordinates on the surface, multiplied by the surface area. 
The formula without $z$ is the integral of the constant $1$ function.
A: This is the ($z$-co-ordinate of the) centre of mass of the hemispherical shell, multiplied by the surface area. Think of the centre of mass of a set of weights $a_i$ at positions $x_i$ on the real axis: that would be
$$ \bar{x} = \frac{\sum_i x_i a_i}{\sum_i a_i}. $$
This is exactly the same, but in higher dimensions and with a surface density replacing discrete points.
A: I see two physical interpretations. Since the hemisphere is symmetric under rotations, its centroid will lie along the $z$-axis, and is given by $$C_z = \frac{\int_S z\,dS}{\int_S dS}.$$
So your first integral, up to a factor of the surface area, is the $z$-coordinate of the centroid. This can be generalized by supposing that the hemisphere has a non-uniform surface density $\sigma$, in which $dS\to \sigma\,dS$ and the integral now is interpreted as the $z$-coordinate of the hemisphere's center of mass.
The second is to interpret both of those integrals is in terms of the flux of a field through the surface. To be precise, the flux of a vector field  $\vec{F}$ through a surface $S$ is defined as $$\phi_F=\int_S \vec{F}\cdot d\vec{S}=\int_S F_n dS$$ where $d\vec{S}=\hat{n}\,dS$ with $\hat{n}$ the local surface normal.
Since the surface normal of a sphere is the radial unit vector $\hat{r}$, one may check that the integrals above are generated by the vector fields $\hat{r}$ and $\hat{z}$ respectively. If one interprets $\vec{F}$ as specifically an electric field, then the two integrals have the interpretation of 'flux due to a unit electric charge at the origin' and 'flux due to a uniform vertical electric field' respectively.
A: Think in this way the surface area function is dependent upon the radius of the hemisphere. 
Take it easy the formula of surface area of a hemisphere is $2*\pi*r(vector)^2$. 
Since r (radius vector)  has z co-ordinates as it is a 3D object so there will be a dot product and the answer will come out to be $2*\pi*(xi+yj+zk).(xi+yj+zk)$ which is equals to $2*\pi*(x^2+y^2+z^2)$. Which can be thought as a general case.
