I agree with wlkingagh.
Given that we know the answer is the same no matter what the diameter of the hole is. Assume the diameter of the whole is 0.
There is no hole, so the remaining volume of the sphere is just the same as the volume of a sphere with a diameter of 6’.

what if the volume of the hole is zero? Then the diameter is 6, radius is 3. 4/3*pi*r^3. r=3. 36pi!! easy peasy!

Similarly, including the cylindrical drilled hole, just alters the limits of integration (draw a picture and it makes sense)
V_sph_with_drilled_hole = ??? r dr d(theta) dz
where now:
r goes from r0 to sqrt(R² – z²),
z goes from -sqrt(R² – r0²) to +sqrt(R² – r0²),
and theta goes from 0 to 2pi.
Doing the integrations you then find
V_sph_with_drilled_hole = (4pi/3) * (R² – r0²)^(3/2)
or V_sph_with_drilled_hole = (4pi/3)*(h/2)³.

Sorry the above comments were formatted poorly, testing
R² = (h/2)² + r0² = 9 + r0²

Similarly, including the cylindrical drilled hole, just alters the limits of integration (draw a picture and it makes sense)
V_sph_with_drilled_hole = ??? r dr d(theta) dz
where now:
r goes from r0 to sqrt(R^{2 – z}2),
z goes from -sqrt(R^{2 – r0}2) to +sqrt(R^{2 – r0}2),
and theta goes from 0 to 2pi.
Doing the integrations you then find
V_sph_with_drilled_hole = (4pi/3) * (R^{2 – r0}2)^{(3/2)
or V_sph_with_drilled_hole = (4pi/3)*(h/2)}3.

Using calculus its pretty straightforward doing the problem in cylindrical coordinates. E.g., calling radius of cylinder r0 and radius of sphere R (where we know that R^{2 = (h/2)}2 + r0^{2 = 9 + r0}2 if the cylinder is of length h), we have for a sphere with no hole:
V_sph = ??? r dr d(theta) dz
where limits of integration are:
r goes from 0 to sqrt(R^2-z2),
z goes from -R to R, and
theta from 0 to 2pi.
Then you quickly find V_sph = (4/3) pi R^3.

I tried using calculus
4/3piR^{3 – [(hpiR}2) + 2pi(integral w/ limits 3 to R-3)(1/x)^2] Problem is I had to guess the function of the decreasing radius of the sphere towards the infinite limit. I think that’s where I got off. Had I that function It’s not that hard.