Calculating the area of part of the surface area $x^2+y^2+z^2=4$ which lies inside the surface $x^2+y^2=2x$

Question

The area of the part of the surface $x^2+y^2+z^2=4$ which lies inside the surface $x^2+y^2=2x$ is equal to $n(\pi-2)$ for an integer n. What is the value of n?

What I know so far:

I get that my answer is $4(\pi-2)$ hence n=4?

Answer 1

One is a sphere of radius $R=2$, and the other is a cylinder of radius $1=R/2$.
Keeping the ratio of the radii, the area intercepted on the sphere shall be proportional to $R^2$ at varying $R$ : thus $n=2^2$