Usually the proof for moment of inertia of a sphere involves taking the element to be a horizontal circular slabs and then integrating it.
I want to use another method to arrive at the same conclusion.
So my idea is, can we consider splitting the circle into small segments of surface area and thinckness?
Something like a structure of an onion.
If so, can you explain how?
If not, can you give the reason why isn't this possible?
The simplest method uses the symmetries of the sphere.
As $ J_{OX}=J_{Oy}=J_{Oz}=J$
we can say that :$$ 3J=\rho \int _{V}(x^{2}+y^{2})\,\mathrm {d} V+\rho \int _{V}(z^{2}+y^{2})\,\mathrm {d} V+\rho \int _{V}(x^{2}+z^{2})\,\mathrm {d} V$$ $$=\rho \int _{V}2(x^{2}+y^{2}+z^{2})\,\mathrm {d} V=2\rho \int _{V}(r^{2})\,\mathrm {d} V$$ where $r$ is the distance from point $M$ to the origin. so :$$ 3J=2\rho \int _{V}(r^{2})\,\mathrm {d} V$$ $$ 3J=2\rho \int _{r}r^{2}\ (\int _{S}\,\mathrm {d} S)\mathrm {d} r=2\rho \int _{r}r^{2}\ \, S\;\mathrm {d} r$$ $$ 3J=2\times 4\pi \rho \int _{r}r^{4}\mathrm {d} r$$ $$ 3J=2\times 4\pi \rho {\frac {R^{5}}{5}}$$ with $$ \rho ={\frac {m}{{\frac {4}{3}}\pi \ R^{3}}}$$ $$ 3J=3{\frac {2mR^{2}}{5}}$$ so: $$ J={\frac {2mR^{2}}{5}}$$
