I will define the metric of $S^{n-1}$ via pullback of the Euclidean metric on ${\mathbb{R}}^{n}$.
To start with we take $n$-dimension Cartesian co-ordinates:
$(x_1,x_2......x_n)$.
The metric here is $g_{ij }= \delta_{ij}$, where $δ$ is the Kronecker delta.
We specify the surface patches of $S^{n-1}$ by the parametrization $f$:
$$x_1=r{\cos{\phi_1}},$$
$$x_p=r{\cos{\phi_p}}{\Pi_{m=1}^{p-1}}{\sin{\phi_{m}}},$$
$$x_n=r{\prod_{m=1}^{n-1}}{\sin{\phi_{m}}},$$
Where $r$ is the radius of the hypersphere and the angles have the usual range.
We see that the pullback of the Euclidean metric $g'_{ab} = (f^*g)_{ab}$ is the metric tensor of the hypersphere. Its components are:
$$g'_{ab} = g_{ij} {\frac{\partial{x_i}}{\partial{\phi_a}}} {\frac{\partial{x_j}}{\partial{\phi_b}}} = {\frac{\partial{x_i}}{\partial{\phi_a}}}{\frac{\partial{x_i}}{\partial{\phi_b}}}$$
We get $2$ cases here:
i) $a>b$ or $b>a$, For these components one obtains a series of terms with alternating signs which vanishes, $g'_{ab}=0$ and thus all off-diagonal components of the tensor vanish.
ii) $a=b$,
$$g'_{11}=1$$
$$g'_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$
where $2\leq a\leq {n-1}$
The determinant is very straightforward to calculate:
$$ \det{(g'_{ab})} = {r^2} \prod_{m=1}^{n-1} g'_{mm}$$
Finally, we can write the metric of the hypersphere as:
$$g' = {r^2} \, d\phi_{1}\otimes d\phi_{1} + {r^2} \sum_{a=2}^{n-1} \left( \prod_{m=1}^{a-1} \sin^2{\phi_{m}} \right) d\phi_{a} \otimes d\phi_{a} $$
Not the simple substitution. The problem changes completely. Now you have all mass in the shell, so is, almost each piece of the sphere farther than before from the axis and it mounts that the farther from the shell before, the more contributing to the moment of inertia now. But, it's true now as before that the tensor is diagonal, so it's only needed an integral about any axis passing by the center. Consider the sphere centered in the origin and calculate the moment of inertia around the $z$ axis. We can do it better in spherical coordinates.
The distance from any point of the shell to the $z$ axis is $r\sin\phi$
and the (constant) density of mass $\rho$
$$I_z=\int_{R}^{R+a}\int_0^\pi\int_0^{2\pi}r^2\sin^2\phi r^2\sin\phi\,\rho\,\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\int_0^{2\pi}\mathrm d\theta\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\int_0^\pi\sin^3\phi\,\mathrm d\phi\mathrm dr=$$
$$=2\pi\rho\int_{R}^{R+a}r^4\left[(1/12)(\cos3\phi-9\cos\phi\right]_0^\pi\mathrm dr=$$
$$=\frac 83\pi\rho\int_{R}^{R+a}r^4\mathrm dr$$
Now, we can apply the condition $a\ll R$. Define $K(x)=\int_{R}^{R+x}Rr^4\mathrm dr$
$$K(x)=\int_{R}^Rr^4\mathrm dr+R^4x+O(x^2)=R^4x+O(x^2)$$
Then, for this shell:
$$I_z\approx\frac 83\pi\rho R^4a$$
Now, the shell has mass $M$ and in a symilar way as for the moment of inertia, we have the volume of the shell is in good approximation $V=4\pi R^2a$ and the mass is $M=4\pi\rho R^2a$. Finally, we can write:
$$I_z=\frac 23MR^2$$
Very different from the simple substitution.
Best Answer
By symmetry the angular momentum is proportional to angular velocity. The only way to have this is when off diagonal elements of the inertia tensor are null.