속이 채워진 직육면체의 회전관성

$$\title{Rotational Inertia of Cuboid}$$ $$ r^2=x^2+ y^2,$$ $$ \rho=\frac{\dd m}{\dd v}=\frac{M}{V}=\frac{M}{X Y H}\taag1$$ $$ \dd v=\dd x \dd y \dd h\taag2 $$ $$ \begin{aligned} \dd m&=\rho\cdot\dd v\\ &=\rho\cdot(\dd x \dd y \dd h)\\ &=\rho\cdot\dd x \dd y \dd h\taag3\\ \end{aligned} $$ $$ \begin{aligned} \dd I&={r}^2\cdot \dd m\\ &=(x^2+ y^2)\cdot(\rho\cdot\dd x \dd y \dd h)\\ &=\rho(x^2+ y^2)\cdot\dd x \dd y \dd h\\ \end{aligned} $$ $$\put L^2=X^2+ Y^2,$$ $$ \begin{aligned} I_{\text{Cuboid}}&=\iiint_V \dd I\\ &=\int_{0}^{H}\int_{-\frac{Y}{2}}^{\frac{Y}{2}}\int_{-\frac{X}{2}}^{\frac{X}{2}} \rho(x^2+ y^2)\cdot\dd x \dd y \dd h\\ &=\rho\int_{-\frac{Y}{2}}^{\frac{Y}{2}}\int_{-\frac{X}{2}}^{\frac{X}{2}} (x^2+ y^2)\cdot\dd x \dd y \cdot\int_{0}^{H}\dd h\\ &=4\rho\int_{0}^{\frac{Y}{2}}\int_{0}^{\frac{X}{2}} (x^2+ y^2)\cdot\dd x \dd y \cdot (H)\\ &=4 \cdot \rho \cdot \int_{0}^{\frac{Y}{2}}\(\frac{X^3}{24}+\frac{X y^2}{2}\) \dd y \cdot H\\ &=4\cdot\(\frac{M}{X Y H}\)\cdot\(\frac{X^3 Y}{48}+\frac{X Y^3}{48}\) \cdot H\\ &=\frac{1}{12}M(X^2+Y^2)\\ &=\frac{1}{12}ML^2\\ \end{aligned} $$