I have this solution this solution here which has radius 2 instead of 4. Hope this might help.
y=(4-x^2)^(1/2) is the outer radius of the surface of revolution
y=1 this is the inner radius of the surface of revolution
Now we need to know the interval of integration...
(4-x^2)^(1/2)=1
4-x^2=1
x^2-3=0, x=±√3
V=p⌠f(o)^2-f(i)^2 dx
V=p⌠4-x^2-1 dx
V=p(3x-x^3/3)
V=(p/3)(9x-x^3), x=[√3, -√3]
V=(p/3)((9√3-3√3)-(-9√3+3√3))
V=(p/3)(6√3+6√3)
V=(p/3)(12√3)
V=4p√3
What has happened is that you have essentially removed a cylindrical tube with convex ends matching the curvature of the outer sphere...it isn't pretty, that's why we needed to use revolutions of solids to figure it out...It would be extremely difficult to figure this out otherwise...
