$I_{remaining} = I_{full} - I_{hole}$
$I_{hole}$ (by parallel axis theorem): $= mr^2/2 + m(R-r)^2$
The difference $= I_{orig} - I_{remaining}$ yields $(m-M)(R+r)^2$ per the question
Answer: $(m-M)(R+r)^2$
$I = \sum m_i r_i^2 = \int r^2\,dm$ (MOI = rotational inertia). Depends on: mass, distribution about axis. Parallel axis theorem: $I = I_{cm} + Md^2$. Perpendicular axis theorem (planar only): $I_z = I_x + I_y$.
$\tau_{net} = I\alpha$ (rotational Newton\'s 2nd law). Angular momentum: $L = I\omega$; $\tau = dL/dt$. If $\tau = 0$: $L = $ const → $I\omega = $ const. Skater spinning: pulls arms in → $I$ decreases → $\omega$ increases (angular momentum conserved). Rotational KE: $KE_{rot} = \frac{1}{2}I\omega^2$. Work by torque: $W = \tau\theta$.
Rolling without slipping: $v_{cm} = R\omega$, $a_{cm} = R\alpha$. Total KE = translational + rotational = $\frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv_{cm}^2(1 + I/mR^2)$. For disc: total KE = $\frac{3}{4}mv^2$. For ring: $mv^2$. For sphere: $\frac{7}{10}mv^2$. Rolling on incline: $a = g\sin\theta/(1 + I/mR^2)$. Sphere accelerates fastest down incline (lowest $I/mR^2$), ring slowest.
Conservation of angular momentum explains: spinning up of pulsars (neutron stars) when stellar collapse reduces $R$ dramatically → $I$ decreases → $\omega$ increases enormously (can reach 700 rotations/second). Same principle for ice skater. Also: Earth\'s rotation is gradually slowing due to tidal friction from Moon → angular momentum transferred from Earth\'s rotation to Moon\'s orbit (Moon slowly moves away from Earth at ~3.8 cm/year).
The parallel axis theorem gives the moment of inertia about any axis parallel to one through the centre of mass: $I = I_{cm} + Md^2$. Note it only works when the reference axis passes through the centre of mass — applying it from any other axis gives a wrong answer. The perpendicular axis theorem, $I_z = I_x + I_y$, applies only to planar laminae, where z is perpendicular to the plane. Using it on a three-dimensional body is a common and costly error.
For a ring about its central axis $I = MR^2$; for a disc, $\frac{1}{2}MR^2$; for a solid sphere, $\frac{2}{5}MR^2$; for a hollow sphere, $\frac{2}{3}MR^2$; and for a thin rod about its centre, $\frac{1}{12}ML^2$, becoming $\frac{1}{3}ML^2$ about one end. Notice the pattern: the further the mass sits from the axis, the larger the coefficient. The rod result also demonstrates the parallel axis theorem directly, since $\frac{1}{12}ML^2 + M(L/2)^2 = \frac{1}{3}ML^2$.
The rotational analogue of Newton's second law is $\tau = I\alpha$, and every linear quantity has a rotational partner: mass becomes moment of inertia, force becomes torque, momentum becomes angular momentum, and $\frac{1}{2}mv^2$ becomes $\frac{1}{2}I\omega^2$. Learning the pairs rather than the individual formulas halves the work, because any linear result can be translated across.
Treating moment of inertia as a fixed property of a body. It depends on the axis chosen. The same rod has three different values about its centre, its end, and a perpendicular axis.
Applying the perpendicular axis theorem to a sphere or cylinder. It is valid only for flat, two-dimensional objects.