# Misconception with the use of Green's functions? [closed]

Problem: Consider a spherical shell of radius R centered on the origin. The top-right quadrant ($$0<θ<90$$ degrees, $$0<φ<90$$ degrees) is at a voltage V. The rest of the sphere is grounded. Determine the potential at the origin.

I am trying to solve this problem using Green's Functions.

For a sphere, $$\phi = \frac{1}{4\pi\epsilon_o}\int\rho G_D d\tau'+\frac{-1}{4\pi}\int\phi(r')\frac{\partial{G_D}}{\partial n'}da'$$

Since charge density is zero everywhere, I know the first term goes to zero. When I find the derivative of $$G_d$$ wrt $$n' = r'$$ (because $$r'$$ points outside the spherical volume) at $$r' = R$$, I get $$\frac{\partial{G_D}}{\partial n'} = \frac{\frac{r^2}{R}-R}{[r^2-2rR\cos(\gamma)+R^2]^\frac{3}{2}}$$

where $$\gamma$$ is the angle between the r and r' vectors. Now I have to take the integral of this on the surface area of the sphere where V is defined as the potential.

However, since I am looking for the potential at the origin, r has to go to zero. This means there is no angle $$\gamma$$ between r and r', so what do I integrate over?

When I plug in $$r = 0$$ into the integral, the potential becomes $$0$$ at the origin. When I assume there is no angle $$\gamma$$ between $$r$$ and $$r'$$ such that $$\gamma = 0$$, then the $$\sin(\gamma)$$ term in $$da' = r^2\sin(\gamma)d\gamma' d\phi'$$ yields zero and just like before the whole thing goes to zero. I do not think this is correct. What am I doing wrong? I must have some misconceptions about the process, but I can't figure out what it is.

sin$$(\gamma)$$ here should be sin$$(\gamma')$$, where $$\gamma'$$ is DIFFERENT (obviously!; I don't know how or why I missed this from the beginning) from the angle in cos$$(\gamma)$$. Also, r in $$da'$$ term be r' = R.