## Integral Considered as a Potential Function

To find
$\int_s \frac{\alpha}{\sqrt{(x-x_0)^2 +(Y-Y_0)^2 +(Z-Z_0)^2}}dS$
, the surface being any surface whatsover with the point
$(x_0 , y_0, z_0 )$
inside it we can consider the integral as a potential function. If we treat the surface as a conducting surface, it will be an equipotential, and the potential inside it will be constant.
If the surface
$S$
$r$
then we can take
$(x_0 , y_0, z_0 )$
to be at the centre of the sphere and translate the integral to the origin.
The integral becomes
$\int^{ \pi}_0 \int^{2 sin \theta \pi}_0 \frac{\alpha}{r} r^2 sin \theta d \theta d \phi dS= 2 \pi \alpha r \int^{\pi} sin \theta d \phi = 2 \pi \alpha r [- cos \theta]^{\pi}_0 = 2\pi \alpha r (-c0s \pi - (-cos 0))= 4 pi\alpha r$