I was trying to derive the rotational kinetic energy of a solid sphere with radius $R$ using the following idea.

Each infinitesimally small shell volume $4\pi r^2 \cdot dr$ at radius $r$ within the sphere has a mass $m$ of $m(r) = \frac{4\pi r^2 \cdot dr}{V} \cdot M$ where $V$ and $M$ are respectively the total volume and total mass of the sphere.

That piece of shell mass is rotating at a velocity of $r\omega$ where $\omega$ is the angular velocity. Such that the rotational kinetic energy of that piece of shellmass, according to $\frac{1}{2}mv^2$, is: $$E_{rot}(r\geq r+dr)=\frac{1}{2} \cdot \frac{4\pi r^2 \cdot dr}{V} \cdot M \cdot (r \cdot w)^2$$ Integrating this equation all up to the total radius $R$ of the sphere would give the total rotational kinetic energy of the sphere. The integration would give: $$E_{rot}=\int_0^R \frac{1}{2} \cdot \frac{4\pi r^2 }{V} \cdot M \cdot (r \cdot w)^2 \cdot dr = \frac{3}{10} M (\omega \cdot R)^2$$ However, the correct rotational kinetic energy formula for a solid sphere has a factor of $\frac{1}{5}$ instead of $\frac{3}{10}$ and integrates over mass increments $dm$ instead of over radius increments $dr$.

Why is my idea for the derivation wrong?


It’s wrong because you assumed that all of the shell is rotating with speed $r\omega$. Only the equator of the shell has that speed. The poles aren’t moving at all. You have to use the distance from the axis of rotation. The spherical coordinate $r$ is the distance from the origin, not the distance from the axis of rotation.

By the way, you are also misusing the word “velocity” when you mean “speed”.

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