Add. Dets.: You don't have to subtract the caps if you don't include them in the first place.
the radius of the sphere, r
the radius of the drill, a
Put the origin of coordinates at the center of the sphere, and the z-axis along the axis of the drill.
After drilling, the remaining solid goes from z = -c to c, where
c^2 = r^2 - a^2
Its cross-section at a plane of constant z is an annulus with:
inner radius a, and
outer radius b = √(r^2 - z^2)
so its cross-sectional area is
A(z) = π(b^2 - a^2) = π(r^2 - a^2 - z^2) = π(c^2 - z^2)
The volume of the solid is then
V = ∫[z=-c,c] A(z)dz
= 2∫[z=0,c] A(z)dz
= 2π ∫[z=0,c] (c^2 - z^2)dz
= 2π (c^2 z - z^3 /3) [z=0,c]
= 2π [c^2 c - c^3 /3]
= 2π (1 - 1/3)c^3
when a=0, c = r, and this is just the volume of the sphere ... √
when a=r, c = 0, V=0 ... √
So for a = 4, r = 9,
c^2 = 81 - 16 = 65
V = (4π/3)c^3 = (4π/3)65^(3/2) = 260π√65 /3 = 2195.1219...
sda, Kdr: Yes, sda, Craig has the wrong formula for the sphere, but you both have the wrong formula for this figure -- If the drill radius were 8 or even 7.35, your method would yield a negative volume!
Kdr: Your cylinder needs a radius other than r, which you've chosen as the radius of the sphere. Call it a, and you have
V = 4π/3 r^3 - π a^2 h
and when you take h = 2r, this becomes
V = 2π r((2/3)r^2 - a^2)
So when a satisfies
2/3 < (a/r)^2 < 1,
you'll get V < 0, when it should be > 0
Hint: a cylinder of height 2*9=18 units cannot lie entirely inside a sphere of diameter=18 units!