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?
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2Can you give more details about your system? You are talking about physics, but you show an equation... What kind of system is it? What are you trying to measure? What is the context? – Martigan Jun 09 '15 at 16:36
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Second one is the surface area. First one, could be anything. – Gappy Hilmore Jun 09 '15 at 16:51
4 Answers
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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.
Berci
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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.
Chappers
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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.
Semiclassical
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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.
