By setting up the integral and integrating, we get
\[R = \frac{\rho a}{\pi b^2} \ln \left| \frac{a+c}{a-c} \right|\]
This expression is valid for all \(c < a\).
For a similarly cut sphere(\(a=b\)),
\[R = \frac{\rho}{\pi a} \ln \left| \frac{a+c}{a-c} \right|\]