속이 채워진 얇은 원반의 회전 관성

$$\title{Rotational Inertia of Solid Disk}$$ $$ \sigma=\frac{\dd m}{\dd a}=\frac{M}{A}=\frac{M}{\pi R^2} \taag1$$ $$ \begin{aligned} l&=r\theta,\\ \dd l&=r \dd \theta\\ \end{aligned} $$ $$ \begin{aligned} \dd a&=\dd r \cdot \dd l\\ &=\dd r \cdot (r \dd\theta)\\ &=r\cdot\dd r \dd\theta\taag2\\ \end{aligned} $$ $$ \begin{aligned} \dd m&=\sigma\cdot\dd a\\ &=\sigma\cdot (r\dd r \dd\theta)\\ &=\sigma r\cdot\dd r \dd\theta\taag3\\ \end{aligned} $$ $$ \begin{aligned} \dd I&=r^2\cdot \dd m\\ &=r^2\cdot \(\sigma r\dd r \dd\theta\)\\ &=\sigma r^3\cdot\dd r \dd\theta\taag4\\ \end{aligned} $$ $$ \begin{aligned} I_{\text{Solid Disk}}&=\oiint_A \dd I\\ &=\int_0^{2\pi}\int_0^{R} \sigma r^3\dd r \dd\theta\\ &=\sigma\cdot\(\int_0^{R} r^3\dd r\)\cdot\(\int_0^{2\pi} \dd\theta\)\\ &=\(\frac{M}{\pi R^2}\)\cdot\(\frac{1}{4}R^4\)\cdot(2\pi)\\ &=\frac{1}{2}MR^2\\ \end{aligned} $$