The problem is the following. There is large body A to which we know the center of gravity coordinates relative to the center of gravity of a shape (that is part of A), specifically a uniform density cylinder, which we call B. Below is a picture of this statement, this is not to scale, and I apologize if the axis look out of place.
The new axis $x',y',z'$ are parallel in their respective planes to the original $x,y,z$ axis. In other words, the new center of gravity merely follows a translational change, and not a rotational one (no yaw, pitch, or roll angle). Now, my job is to calculate the Inertia Tensor of the body A, knowing the inertia tensor without the addition of this body B, but also knowing the center of gravity of A including body B, in the reference frame $x'y'z'$.
This seems fairly straightforward, but I am having some doubts. Here is my current work.
$$I_{xx}=I_{xx_A}+I_{xx_B}+m_B(d_x)^2$$
$$I_{yy}=I_{yy_A}+I_{yy_B}+m_B(d_y)^2$$
$$I_{zz}=I_{zz_A}+I_{zz_B}+m_B(d_z)^2$$
To simplify things, let us assume that the distance $d_y$ is zero. This means the cylinder is offset from the center of gravity of the entire larger body A in only one plane. This also means our inertia tensor will be axisymmetric but will also be a diagonal (we also assume that the body A has a diagonal tensor with and without B).
The result is then:
$$I_{xx}=I_{xx_A}+I_{xx_B}+m_B(d_x)^2$$
$$I_{yy}=I_{yy_A}+I_{yy_B}$$
$$I_{zz}=I_{zz_A}+I_{zz_B}+m_B(d_z)^2$$
And our inertia tensor is $[I_{xx}\ 0\ 0;0\ I_{yy}\ 0;0\ 0\ I_{zz}]$
Background:
The purpose of doing this is for determining the moment of inertia of a launch vehicle. This cylinder approximates the shapes of one of the engines. To me, the problem seems fairly straightforward. However, when I look at an example in my book, they seem to be doing a completely different calculation. Speficially, the $md^2$ part is written as $m*(l-lx-x_{cg})$ where $m$ is the mass of the entire launch vehicle ($m_A$ in this example), $lx$ is the height of the center of gravity location relative to the bottom most part of the launcher (the part that touches the ground, in this case the bottom of our cylinder), and $x_{cg}$ is the center of gravity of the entire launch vehicle (which we know).
The cylinder $c$ has a moment of inertia w.r.t. its centre of gravity $G$ $$I_c^{(G)} = \frac{m}{12} \begin{bmatrix} 6r^2 & 0 & 0\\ 0 & 3r^2+h^2 & 0 \\ 0 & 0 & 3r^3+h^2 \end{bmatrix}.$$ (Notice, however, that textbooks usually define x and z in reverse.)
Apply the parallel axis theorem to find the moment of inertia of the cylinder w.r.t the axes of entity $A$ $$I_c^{(A)} = I_c^{(G)} + m \operatorname{veck}^{-T}\!\!\left(r^{GA}\right) \operatorname{veck}^{-1}\!\!\left(r^{GA}\right)$$ with the inverse vectorise skew operator $$\operatorname{veck}^{-1} X = \operatorname{veck}^{-1}\left(\left[x\quad y\quad z\right]^T\right) = \begin{bmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{bmatrix}.$$
Inserting the distance between $G$ and $A$ $$r^{GA} = \begin{bmatrix} r_x \\ 0 \\ r_z \end{bmatrix},$$ you obtain $$I_c^{(A)} = I_c^{(G)} + m \begin{bmatrix} r^2_z & 0 &-r_x r_z\\ 0 & r^2_x + r^2_z & 0 \\ -r_x r_z & 0 & r^2_x \end{bmatrix}.$$