자연계열 2번

[문제 2] 밀도($\displaystyle \rho$)가 상수이므로, 용기에 가득 담긴 물의 질량은$\displaystyle M=\int_0^a\rho A(z)\,dz=\rho\int_0^a A(z)\,dz=\frac{2\pi}3\rho(a^3-2b^3)$이다. 한편$\displaystyle \begin{aligned}\int_0^a zA(z)\,dz&=\int_0^{2b}zA(z)\,dz+\int_{2b}^a zA(z)\,dz\\&=\int_0^{2b}2\pi(a-b)z^2\,dz+\int_{2b}^a\pi(2az^2-z^3)\,dz\\&=2\pi a\int_0^a z^2\,dz-2\pi b\int_0^{2b}z^2\,dz-\pi\int_{2b}^a z^3\,dz\\&=\frac{2\pi}3a^4-\frac{16\pi}3b^4-\frac\pi4a^4+4\pi b^4\\&=\frac{5\pi}{12}a^4-\frac{4\pi}3b^4\end{aligned}$이므로 $\displaystyle \int_0^a\rho zA(z)\,dz=\rho\frac\pi{12}(5a^4-16b^4)$이다. 따라서 질량중심의 $\displaystyle z$좌표 $\displaystyle \bar z$는$\displaystyle \bar z=\frac{\int_0^a\rho zA(z)\,dz}{\int_0^a\rho A(z)\,dz}=\frac{5a^4-16b^4}{8(a^3-2b^3)}$이다.