That looks lie a homework style question. So here the necessary hints.
The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem
Edit
While I think this is not a calculus help site, I will try to give an almost complete solution.
The dimensions are defined in the image. Note that this is a general solution that covers basically all your cases. For $A=B=0$ it is solid and hollow otherwise. For $B=D$ and $A=C$ it is a (hollow) cylinder. It is simple to verify that for those cases the following results coincide with the special case formulae.
The volume is given by
$$V= 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r$$
where
$$r_\mathrm i(z)= A+(C-A)\frac z h \quad r_\mathrm o(z)= B+(D-B)\frac z h$$
This results in
$$V= \frac {\pi}{ 3} \left[(B^2+BC+D^2)-(A^2+AC+C^2)\right]$$
One can see easily that for $A=B=C=0$ and $D=R$ this gives the volume of a simple cone. The centre of mass is at:
$$\begin{eqnarray}
z_\mathrm{cm}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r z\\
&=&\frac h 4 \frac{(B^2+2BD+3D^2)-(A^2+2AC+3C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$Again we see that in the cone limit this gives the known result of $z_\mathrm{cm}=3 h/4$. Also note that this general formula also allows to put the tip at the top, i.e. only $B\neq 0$ in which case one sees easily that the centre of mass is at $h/4$.
Now for $I_{xx}=I_{yy}$ and $I_{zz}$ (Note, I do this in units of the body's mass, so I divide by the volume, assuming constant density $\rho$):
$$\begin{eqnarray}
I_{zz}&=&\frac 1 V 2 \pi \int_0^h \mathrm d z \int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (x^2+y^2)\\
&=& \frac 3 {10} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$
which is in the above mentioned limit $3/10 R^2$. Finally, we have
$$\begin{eqnarray}
I_{xx}&=&\frac 1 V \int_0^h \mathrm d z \int_0^{2 \pi} \mathrm d \varphi\int_{r_\mathrm i(z)}^{r_\mathrm o(z)} r \mathrm d r (z^2+y^2)\\
&=& \frac 3 {30} h^2 \frac{(B^2+3BD+6D^2)-(A^2+3AC+6C^2)}{(B^2+BD+D^2)-(A^2+AC+C^2)}+\\
&&+\frac 3 {20} \frac{(B^4+B^3D+ B^2D^2+BD^3+D^4)-(A^4+A^3C+A^2C^2+AC^3+C^4)}{(B^2+BD+D^2)-(A^2+AC+C^2)}
\end{eqnarray}$$
which again gives the well known factors $3/5 h^2$ and $3/20 R^2$ in case of $A=B=C=0$ and $D=R$.
Eventually, we need to check the moment in the centre of mass.
Now this I will leave open for the OP. Just consider this:
This calculation is obviously not in the centre of mass.$I_{zz}$ should not change as we shift along this axes. We also know that having the centre of mass solution, going to the coordinate system we have, is done by using the parallel axis theorem, i.e.
$$I_{xx}=I_{xx,\mathrm{cm}}+z_\mathrm{shift}^2$$
(again in units of $m$).
You may, as an alternative, reformulate the last integral.
TL;DR This question is actually more related to mathematics than physics. Here I give just some basic guidelines which can help you in most textbook problems of finding moment of inertia. Please note that finding moment of inertia is not difficult, but it takes practice! In the most complicated cases we use specialized software to numerically compute moment of inertia.
Below you can find three examples on how to calculate moment of inertia for a cube and a cylinder. Make sure to understand these examples before you move on to more complex problems (objects). For anything more detailed than this, consult some book on calculus on how to deal with double and triple integrals.
Moment of inertia in general case
The idea is simple - imagine an object is composed of many many infinitesimally small particles of mass $m_i$ at distance $r_i$ from the axis of rotation. To find moment of inertia you just sum distance squared times mass of each particle
$$I = \sum_{i} r_i^2 m_i$$
The above equation works well in discrete case, but in continuous case the sum becomes integral
$$\boxed{I = \iiint_\mathcal{V} r^2 dm} \tag 1$$
where $\mathcal{V}$ is the volume that object occupies in 3D. Now the only thing you have to do is to represent the infinitesimally small mass $dm$ as a function of distance $r$. For objects whose density $\rho$ is constant (homogeneous objects), the mass can be expressed as
$$dm = \rho \cdot dV$$
where $dV$ is infinitesimally small volume. The integral in Eq. (1) now becomes
$$\boxed{I = \rho \iiint_\mathcal{V} r^2 dV} \tag 2$$
Moreover, if object's cross section does not change along the axis of rotation, the above equation can be further simplified:
$$dm = \rho \cdot dV = \rho \cdot (h \cdot dA) = \sigma \cdot dA$$
where $h$ is object length along axis of rotation, and $\sigma$ is mass-area density of the object. The above simplification converts triple integral (volume) to double integral (area), which is much easier to depict. The integral in Eq. (2) now becomes
$$\boxed{I = \sigma \iint_\mathcal{A} r^2 dA} \tag 3$$
where $\mathcal{A}$ is the area that object occupies in 2D. The double integrals are solved either in Cartesian coordinates or polar coordinates, whichever is more suitable for a specific problem. Except for these two, the triple integrals can also be solved in spherical coordinates, but I will focus here only on double integrals.
Cartesian and polar coordinates
In all cases discussed here we assume that the axis of rotation goes through the origin. You are allowed to place the origin wherever it is most convenient to solve a particular problem (integral)!
In Cartesian coordinates, the infinitesimally small area and distance from origin are defined as
$$dA = dx \cdot dy \qquad \text{and} \qquad r^2 = x^2 + y^2$$
and the Eq. (3) becomes
$$\boxed{I = \sigma \iint_\mathcal{A} (x^2 + y^2) dx dy} \tag 4$$
In polar coordinates, the infinitesimally small area is defined as
$$dA = (r+dr)^2 \frac{d\theta}{2} - r^2 \frac{d\theta}{2} = (2rdr + (dr)^2) \frac{d\theta}{2} \approx r dr d\theta$$
where $(dr)^2$ is neglected being much (much!) smaller than $2rdr$. With this Eq. (3) becomes
$$\boxed{I = \sigma \iint_\mathcal{A} r^2 \cdot r dr d\theta} \tag 5$$
It must be noted that Eq. (4) and Eq. (5) are only valid if:
- object is homogeneous, i.e. its density $\rho$ is constant, and
- object's cross section does not change along the axis of rotation!
In most textbook problems the above two requirements are satisfied. If not, you have to use Eq. (2) for homogeneous objects or Eq. (1) for non-homogeneous objects.
Example 1: Cube with all sides equal to $L$
Assume that the axis of rotation goes through center of mass and is perpendicular to two faces. Moment of inertia is
$$I = \sigma \int_{-L/2}^{L/2} \int_{-L/2}^{L/2} (x^2 + y^2) dx dy = \frac{1}{6} M L^2$$
where $M = \sigma L^2$ is the total mass of the cube. Integration is done in Cartesian coordinates, and the axis of rotation is placed at the origin.
Example 2: Cylinder with radius $R$
Assume that the axis of rotation goes through center of mass and is perpendicular to cylinder's base. Moment of inertia is:
$$I = \sigma \int_{0}^{R} \int_{0}^{2\pi} r^3 dr d\theta = \frac{1}{2} M R^2$$
where $M = \sigma R^2 \pi$ is the total mass of the cylinder. Integration is done in polar coordinates, and the axis of rotation is placed at the origin.
Parallel-axis theorem
The parallel axis theorem, also known as Huygens-Steiner theorem, relates moment of inertia of an object about axis through center of mass to moment of inertia about arbitrary parallel axis
$$\boxed{I = I_\text{cm} + M d^2} \tag 6$$
where $M$ is total mass of the object and $d$ is distance between the two parallel axis.
Example 3: Cube rotating about edge
In Example 1 we have found moment of inertia for cube that rotates about axis through center of mass. When axis goes along cube edge, moment of inertia is
$$I = \sigma \int_{0}^{L} \int_{0}^{L} (x^2 + y^2) dx dy = \frac{2}{3} M L^2$$
We could have calculated this using the parallel-axis theorem. The axis of rotation is at the distance $d = \frac{\sqrt{2}}{2} L$ from axis that goes through center of mass, hence
$$I = \frac{1}{6} M L^2 + M \bigl( \frac{\sqrt{2}}{2} L \bigr)^2 = \frac{2}{3} M L^2$$
Best Answer
The formulae from the NASA document and Wikipedia are simply in different frames: the ones in Wikipedia assume coordinate frame whose axes pass through the center of mass whereas the NASA document assumes arbitrary cartesian coordinate frame.
Let capital Xi, Yi and Zi denote coordinates of i-th atom in an arbitrary Cartesian coordinate frame O and xi, yi and zi denote respective coordinates in the frame Ocom translated to the center of mass of the molecule.
The coordinates of the center of mass of the molecule in the O frame are:
\begin{equation} X_{com} = \frac{1}{M}\sum_{i} m_i X_i \\ Y_{com} = \frac{1}{M}\sum_{i} m_i Y_i \\ Z_{com} = \frac{1}{M}\sum_{i} m_i Z_i \end{equation}
Transformation from O to Ocom is simply translation by [-Xcom, -Ycom, -Zcom]:
\begin{equation} x_i = X_i - X_{com} = X_i - \frac{1}{M}\sum_{i} m_i X_i\\ y_i = Y_i - Y_{com} = Y_i - \frac{1}{M}\sum_{i} m_i Y_i\\ z_i = Z_i - Z_{com} = Z_i - \frac{1}{M}\sum_{i} m_i Z_i \end{equation}
Now, take for example the formula for Ixx in the O frame:
\begin{equation} I_{xx} = \sum_{i} m_i (y_i^2 + z_i^2) \end{equation}
(Note that the two-term sum in the parenthesis is simply the Euclidean distance between the i-th atom and the x axis, exactly as one would expect.)
Substituting the formulae for yi and zi above we have
\begin{equation} I_{xx} = \sum_{i} m_i [(Y_i - Y_{com})^2 + (Z_i - Z_{com})^2] \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) + \sum_{i} m_i (Y_{com}^2 + Z_{com}^2) - 2 \sum_{i} m_i (Y_i Y_{com} + Z_i Z_{com}) \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) + M (Y_{com}^2 + Z_{com}^2) - 2 M (Y_{com}^2 + Z_{com}^2) \\ I_{xx} = \sum_{i} m_i (Y_i^2 + Z_i^2) - \frac{1}{M}(\sum_{i} m_i Y_i)^2 - \frac{1}{M}(\sum_{i} m_i Z_i)^2 \end{equation}
which is exactly the formula for Ixx in the NASA document. Analogously for the other formulae.
Interestingly, the sum of the last two terms in the final formula is exactly what one would expect from the parallel axis theorem.
The conclusion is that you should use whichever formulae fit your coordinate frame. If the center of your coordinate frame coincides with the center of mass of the molecule, then you can use the simpler formulae from Wikipedia. Otherwise you should use those from the NASA document.