= Solution
Put $z=x_3$. The body is a <solid of revolution> with disk radius
$$
R(z)=z\sqrt{1-\frac zh},
\qquad 0\leq z\leq h.
$$
Its volume and uniform <mass density> are
$$
V=\pi\int_0^hR(z)^2\,dz
=\pi\int_0^h\left(z^2-\frac{z^3}{h}\right)dz
=\frac{\pi h^3}{12},
\qquad
\rho=\frac{12M}{\pi h^3}.
$$
By the continuous <center of mass> formula, the distance of the center of mass from the pointed origin is
$$
l=\frac{\rho\pi}{M}\int_0^h zR(z)^2\,dz
=\frac{12}{h^3}\left(\frac{h^4}{4}-\frac{h^4}{5}\right)
=\boxed{\frac{3h}{5}}.
$$
For a disk of radius $R$, $\int r^2\,dA=\pi R^4/2$. Therefore the axial moment is
$$
I_3=\frac{\rho\pi}{2}\int_0^hR(z)^4\,dz
=\frac{\rho\pi}{2}h^5\int_0^1u^4(1-u)^2\,du
=\boxed{\frac{2}{35}Mh^2}.
$$
For the transverse $x_1$-axis, $\int x_2^2\,dA=\pi R^4/4$, while every slice also contributes $z^2$ times its mass. Hence
$$
\begin{aligned}
I_1
&=\rho\pi\int_0^h\left(\frac{R(z)^4}{4}+z^2R(z)^2\right)dz\\
&=\rho\pi h^5\left(\frac1{4\cdot105}+\frac1{30}\right)
=\boxed{\frac37Mh^2}.
\end{aligned}
$$
By axial symmetry $I_2=I_1$. These values agree with the <moment of inertia of a pointed cubic-profile solid of revolution>.
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