# Question on the proof of Moment of Inertia for a solid sphere

I was trying to prove myself the moment of inertia of a solid sphere. Here's it,

Let us consider a sphere in $$3$$-dimensions with radius $$R$$ and mass $$M$$. Moment of inertia is the product of mass with its perpendicular distance from the axis of rotation.

Let us consider a hollow sphere inside our original sphere with radius $$r\lt R$$ with infinitesimal thickness and infinitesimal mass $$dm$$. Assuming that the body is uniform I can state that, $$\frac{M}{4\pi R^{2}}=\frac{dm}{dV}$$

$$\frac{dV}{dr}=4\pi r^{2}$$ $$dV=4\pi r^{2} dr$$ Putting it back in the equation we get,

$$\frac{M}{4\pi R^{2}}=\frac{dm}{4\pi r^{2}dr}$$ $$dm=\frac{3Mr^{2}dr}{R^{3}}$$

From the definition of moment of Inertia we know that, $$I=\int_{}^{}r^{2}dm$$ $$I=\int_{0}^{R}\frac{3Mr^{4}}{R^{3}}dr$$

Solving it we get, $$I=\frac{3}{5}MR^{2}$$

But textbook says that the moment of Inertia of solid sphere is $$\frac{2}{5}MR^{2}$$

My question:-

$$(1)$$- Obviously there is some mistake in the proof, but I don't understand where the mistake is?

You've misunderstood what $$r$$ means in $$I=\int r^2\,\mathrm dm$$. It refers not to the $$r$$ from the spherical polar coordinates with respect to which the solid is the locus $$r\le R$$, but the perpendicular distance of an arbitrary point from the axis about which the sphere's moment of inertia is computed. Without loss of generality we can assume the system of spherical polar coordinates is chosen so that, in terms of that system's radius $$r$$, the distance in question is $$r\sin\theta$$. We therefore need to be more careful than getting $$\mathrm dm$$ for a constant-$$r$$ spherical shell. Since $$\mathrm dm=\frac{3M}{4\pi R^3}r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi$$, the result is a triple integral,$$\frac{3M}{4\pi R^3}\int_0^Rr^4\,\mathrm dr\int_0^\pi\sin^3\theta\,\mathrm d\theta\int_0^{2\pi}\mathrm d\phi,$$which you can verify is $$\frac25MR^2$$.