Sphere moment of inertia using disk method

Question

Need to solve the moment of inertia of a sphere definined as the integral $\int\left(r^2\right)dm$ it can be rewritten as $\int\left(f(x)^2\right)pdV$ where dm= pdV and p represents the mass density of a sphere and $f(x)= sqrt(R^2-x^2)$ . The integral simplifies as $2p\pi\int_0^R\left(R^2-x^2)^2\right)dx$. We find that$\int_0^R\left(R^2-x^2)^2\right)dx$ is equal to $8/15R^5$ if we multiply this by $2p\pi$ we obtain a moment of inertia equal to $12/15mR^2$ which is twice the moment of inertia of a sphere. The problem comes from doubling the integral by transforming the limits -R to R to $2\int_0^R$

Answer 1

The function to integrate is an even function ($f(-x)=f(x)$, or "symmetric about x-axis"), so this statement is justified. See this for a proof if you aren't convinced; I'd just say look at the graph of a function symmetric about the x-axis (e.g. $|x|$, $x^2$) and see for yourself.

As an unrelated note, please try to be more clear in future posts. Ask, perhaps, solely about where the problem comes from, instead of including all of the background about the MOI of a sphere.