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.

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.

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(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.