【構造力学】断面二次モーメントについての話

f_1とf_2の説明
$$\begin{eqnarray} I_x&=&\int y^2\mathrm{d}A\\ &=&\int_a^b y^2\big(f_1(y)-f_2(y)\big)\mathrm{d}y\\ &=&\int_a^b y^2 \int_{f_2(y)}^{f_1(y)}\mathrm{d}x\mathrm{d}y\\ &=&\iint_D y^2\mathrm{d}D \end{eqnarray}$$
故に，
$$\begin{eqnarray} I_x+I_y&=&\iint_D(x^2+y^2)\mathrm{d}D\\ &=&\int_0^{2\pi}\int_0^{r(\theta)}r^3\mathrm{d}r\mathrm{d}\theta\quad{(x\mapsto r\cos\theta,y\mapsto r\sin\theta)}\\ &=&\int_0^{2\pi}\frac{1}{4}(r(\theta))^4\mathrm{d}\theta \end{eqnarray}$$
$r(\theta+2\pi)=r(\theta)$とし，$XY$座標が$xy$座標に対し$\alpha$だけ傾いていて，
$$\begin{eqnarray} I_x+I_y-I_X-I_Y&=&\int_0^{2\pi}\frac{1}{4}(r(\theta))^4\mathrm{d}\theta-\int_\alpha^{2\pi+\alpha}\frac{1}{4}(r(\theta))^4\mathrm{d}\theta\\ &=&\int_0^{\alpha}\frac{1}{4}(r(\theta))^4\mathrm{d}\theta-\int_{2\pi}^{2\pi+\alpha}\frac{1}{4}(r(\theta))^4\mathrm{d}\theta\\ &=&0\quad(\text{第二項を}2\pi+\theta\mapsto\theta\text{と置換}) \end{eqnarray}$$