In my mechanics textbook there is a derivation of the moment of inertia of a disc of mass $m$ and radius $r$ about an axis through its centre and perpendicular to its plane surface, which goes something like this:
The mass per unit area is $\dfrac{m}{\pi r^2}$. Dividing the disc into concentric rings, the ring has inner radius $x$ and outer radius $x+\delta x$ and so its area is $\pi(x+\delta x)^2-\pi x^2\approx 2\pi x\delta x$. So the moment of inertia of this ring is $\dfrac{m}{\pi r^2}(2\pi x\delta x)(x^2)=\dfrac{2mx^3}{r^2}\delta x$ so the moment of inertia of the whole disc is $\displaystyle \lim_{\delta x\to0}\sum_{i=0}^n\dfrac{2mx_i^3}{r^2}\delta x=\int_0^r\dfrac{2mx^3}{r^2}\,\mathrm{d}x=\frac{1}{2}mr^2$.
But the step $\pi(x+\delta x)^2-\pi x^2\approx 2\pi x\delta x$ is hand-wavy. How do I know this is a valid approximation?
Best Answer
The real thing is the constant density $\rho = \frac{m}{R^2\pi}$. What you have to do is to compute the integral $$ \int_0^R r^2 dm = \int_0^R r^2 \rho dA = \int_0^R r^2 \rho 2\pi r dr = 2\pi \rho [\frac{1}{4}r^4]_0^R= 2\pi \rho \frac{1}{4}R^4 = \frac{1}{2} R^2 m $$ the hand-wavy part is just the standard way to avoid integration by handy-wavy-this-is-a-riemann-sum-approximation of the integral