Surface Integral Considered as a Potential Function 2

To find  
\[\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2}}dS\]
, the surface being a sphere not containing the point  
\[(x_0 , y_0 , z_0)\]
  we can consider the integral as a potential function. As long as the distribution of charge over the surface is spherically symmetrical, we can treat the charge distribution over the surface as a point charge at the center of the sphere.
Continuing the potential analygy, take the charge per unit area as 1. If the radius of the sphere is  
  then the charge equals the surface area equals  
\[4 \pi r^2\]
\[\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2)}}dS = \frac{4 \pi r^2}{\sqrt{x_0^2 +y_0^2+z_0^2}} \]

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