How to calculate the inertia tensor of a spherical cap? In this question, an attempt is made at calculating the diagonal elements of the inertia tensor of a homogeneous spherical cap, where the $z$-axis is the symmetry axis. The mass moment of inertia about the $z$-axis is expressed by:
$$M_{zz}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r^3 dr dz d\theta$$
where $\rho$ is the density, given by:
$$\rho = \frac{m}{\frac{1}{3} π h^2 (3 R - h)}$$
$R$ is the radius of the sphere, and $h$ and $m$ are the height and mass of the cap. The triple integral equation solves as:
$$M_{zz}=\frac{mh}{10(3R - h)}(3h^2 - 15hR + 20R^2)$$
which is correct. However, the following expression is given for the mass moment of inertia about the $x$-axis or $y$-axis (in the accepted answer to this question):
$$M_{xx}=M_{yy}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r(r^2 \cos^2\theta+z^2) dr dz d\theta$$
which solves as (see WolframAlpha):
$$M_{xx}=M_{yy}=\frac{m}{20(3R - h)}(-9h^3 + 45h^2R - 80hR^2 + 60R^3)$$
This is seemingly not correct. If we plug in $R=5$, $h=2$ and $m=4.28\times10^{5}$, we get:
$$I=\begin{bmatrix}7.1246&0&0\\0&7.1246&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$
According to CATIA's "Measure Inertia"-function, this should be:
$$I=\begin{bmatrix}1.2896&0&0\\0&1.2896&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$
For reference (don't mind the definition of the axes here):

The question is:
What is the correct expression for the mass moment of inertia about the principal $x$-axis/$y$-axis (red/green lines in reference figure)? 
The geometric centroid (the origin of the axes system) for a spherical cap is given by:
$$ z=\frac{3(2R-h)^2}{4(3R-h)} $$
Edit:
Thanks to probably_someone, the answer can be derived as:
$$M_{xx}=M_{yy}=\frac{mh}{80(h-3R)^3}(-9h^3 + 72h^2R - 220hR^2 + 240R^3)$$
 A: The moment of inertia of an object is only defined relative to a particular choice of axes of rotation. You are using a different choice of $x$ and $y$ axes than the original question.
In the original question, the $x$ and $y$ axes passed through the center of the whole sphere, meaning they passed through a point on the $z$-axis that is below the bottom of the cap. CATIA is calculating the moments of inertia for $x$ and $y$ axes that pass through the center of mass of the cap. Fortunately, there is an easy way to convert between these two choices: the parallel axis theorem. The theorem states:

Suppose a body of mass $m$ has moment of inertia $I_0$ about an axis passing through its center of mass. Then the moment of inertia $I$ about another axis, parallel to the first and displaced a distance $d$ from the center of mass, is given by: $$I=I_0+md^2$$

Applying this to our situation, it's clear that the CATIA calculation gives you $I_0$, and we already have $m$. Since you defined the center of the whole sphere as the origin, the distance $d$ that the axis should be displaced is equal to the $z$-coordinate of the center of mass, namely:
$$d=\frac{3(2R-h)^2}{4(3R-h)}$$
So the moment of inertia about $x$ and $y$ axes passing through the center of the sphere will be:
$$I=I_0+m\frac{9(2R-h)^4}{16(3R-h)^2}$$
Plugging in $m=4.28\times 10^5$, $R=5$, $h=2$, and $I_0=1.2896\times 10^6$, we get $I=7.1246\times 10^6$, which perfectly agrees with the calculation that you did using the other question.
