[Math] Moment of inertia of hemisphere

I want to find the moment of inertia of the hemisphere shown in the picture about the $O$ axis.

I am getting different results depending on the approach.

1) The moment of inertia of the hemisphere about an axis parallel to $O$ passing through its Center of Mass (CoM) is

$
I_{CoM} = \frac{83}{320}ma^2
$

wherein $m$ is the mass of the hemisphere.

The distance from the $O$ axis to its CoM is $\frac{5}{8}a$.
Therefore, from the parallel axis theorem,

$
I_O = \frac{83}{320}ma^2 + m(\frac{5}{8}a)^2=\frac{13}{20}ma^2
$

2) The moment of inertia of the hemisphere about an axis parallel to $O$ passing through $a$ (the "base" of the hemisphere) is

$
I_{CoM} = \frac{2}{5}ma^2
$

Therefore, from the parallel axis theorem,

$
I_O = \frac{2}{5}ma^2 + ma^2=\frac{7}{5}ma^2
$

3) The general expression for the moment of inertia is:

$
I_O = \int_m{r^2dm} = \int_m{(x^2+y^2)\cdot\rho dV} = \int_m{(x^2+y^2)\cdot\rho \cdot (\pi \cdot y^2 \cdot dx)}
$

Substituting $y^2=a^2-(x-a)^2$, we get

$
I_O = \int_0^a{(x^2+(a^2-(x-a)^2))\cdot\rho \cdot (\pi \cdot (a^2-(x-a)^2) \cdot dx)} = \frac{5}{6} \pi \rho a^5
$

The mass of the hemisphere is $m=\frac{2}{3}\pi \rho a^3$, so the expression above can be written in terms of the mass $m$ as

$
I_O = \frac{5}{6} \frac{3}{2} \left( \frac{2}{3} \pi \rho a^3 \right) a^2 = \frac{15}{12}ma^2 = \frac{5}{4}ma^2
$

4) Since the differential volume in (3) was $dV=\pi y^2 dx$, maybe the general expression reduces general for the moment of inertia reduces to:

$
I_O = \int_m{r^2dm} = \int_m{x^2 \cdot\rho dV} = \int_m{x^2 \cdot\rho \cdot (\pi \cdot y^2 \cdot dx)}
$

Which, after substituting $y^2=a^2-(x-a)^2$, results in:

$
I_O = \int_0^a{x^2 \cdot\rho \cdot (\pi \cdot (a^2-(x-a)^2) \cdot dx)} = \frac{3}{10} \pi \rho a^5 = \frac{9}{20}ma^2
$

In summary, four different approaches and four different results. What am I missing here?