東京工業大学前期 1991年第3問 (解答解説)

【解答】

円の中心をとおいてとする（）．この様子を図に示す．

この時の座標は

だからでの円の接線は

と書ける．これらがそれぞれを通るからの方程式に代入して

である．

題意の回転体の体積をとする．この回転体は座標の小さい方から順に，線分を回転させた円錐，間の弧を回転させた部分，線分を回転させた円錐の3つの部分から成る．

\begin is only supported in math mode\begin{align} V(a) &= \left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \node[anchor=east, font=\scriptsize] at (0,0.4) {B}; \draw (0,0.4) -- (1.5,0); \draw (0,0.4) -- (1.5,0.8); \draw (1.5,0.4) ellipse (0.15 and 0.4); \node[anchor=south, font=\scriptsize, yshift=2pt] at (1.5,0.8) {D}; \end{tikzpicture}}\right) + \left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \draw (0,0) -- (1.2,0); \draw (0,0.8) -- (1.2,0.8); \draw (0,0) -- (0,0.8); \draw (1.2,0) -- (1.2,0.8); \draw (0,0.8) to[out=45, in=135] (1.2,0.8); \draw[->] (0.6, -0.1) arc (200:340:0.1); \draw[->] (0.6, -0.1) arc (160:20:0.1); \node[anchor=south east, font=\scriptsize, xshift=2pt] at (0,0.8) {D}; \node[anchor=south west, font=\scriptsize, xshift=-2pt] at (1.2,0.8) {C}; \end{tikzpicture}}\right) + \left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \node[anchor=west, font=\scriptsize] at (1.5,0.4) {A}; \draw (1.5,0.4) -- (0,0); \draw (1.5,0.4) -- (0,0.8); \draw (0,0.4) ellipse (0.15 and 0.4); \node[anchor=south, font=\scriptsize, yshift=2pt] at (0,0.8) {C}; \end{tikzpicture}}\right) \end{align}

各項を個別に計算しよう．わかりやすさのためにから軸におろした垂線の足をとおくと(式1)より

である．は底面の半径高さの円錐であって，は底面の半径高さの円錐であるから(式3)にも注意して

\begin is only supported in math mode\begin{align} \left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \node[anchor=east, font=\scriptsize] at (0,0.4) {B}; \draw (0,0.4) -- (1.5,0); \draw (0,0.4) -- (1.5,0.8); \draw (1.5,0.4) ellipse (0.15 and 0.4); \node[anchor=south, font=\scriptsize, yshift=2pt] at (1.5,0.8) {D}; \end{tikzpicture}}\right)&=\dfrac{1}{3}\left(\pi|DD'|^2\right)\cdot |D'B| \\&=\dfrac{1}{3}\cdot(\pi\sin^2\beta)\cdot(a-\cos\beta+2) \\&=\dfrac{\pi}{3}\left(1-\left(\dfrac{1}{2+a}\right)^2\right)\left(a+2-\dfrac{1}{2+a}\right)\quad(\because\text{(式3)}) \\&=\dfrac{\pi}{3}\cdot\dfrac{\{(2+a)^2-1\}^2}{(2+a)^3}\\\left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \node[anchor=west, font=\scriptsize] at (1.5,0.4) {A}; \draw (1.5,0.4) -- (0,0); \draw (1.5,0.4) -- (0,0.8); \draw (0,0.4) ellipse (0.15 and 0.4); \node[anchor=south, font=\scriptsize, yshift=2pt] at (0,0.8) {C}; \end{tikzpicture}}\right)&=\dfrac{1}{3}\left(\pi|CC'|^2\right)\cdot |C'A| \\&=\dfrac{1}{3}(\pi\sin^2\alpha)(2-a-\cos\alpha) \\&=\dfrac{\pi}{3}\left(1-\left(\dfrac{1}{2-a}\right)^2\right)\left(2-a-\dfrac{1}{2-a}\right)\quad(\because\text{(式3)}) \\&=\dfrac{\pi}{3}\cdot\dfrac{\{(2-a)^2-1\}^2}{(2-a)^3} \end{align}

である．最後に円弧の回転体部分は

\begin is only supported in math mode\begin{align} \left(\text{ \begin{tikzpicture}[baseline={(0,0.1)}, scale=0.8, >=stealth] \draw (0,0) -- (1.2,0); \draw (0,0.8) -- (1.2,0.8); \draw (0,0) -- (0,0.8); \draw (1.2,0) -- (1.2,0.8); \draw (0,0.8) to[out=45, in=135] (1.2,0.8); \draw[->] (0.6, -0.1) arc (200:340:0.1); \draw[->] (0.6, -0.1) arc (160:20:0.1); \node[anchor=south east, font=\scriptsize, xshift=2pt] at (0,0.8) {D}; \node[anchor=south west, font=\scriptsize, xshift=-2pt] at (1.2,0.8) {C}; \end{tikzpicture}}\right)&=\pi\int_{a-\cos\beta}^{a+\cos\alpha}\{1-(x-a)^2\}dx \\&=\pi\int_{-\cos\beta}^{\cos\alpha}(1-x^2)dx \\&=\pi\left[x-\dfrac{1}{3}x^3\right]_{-\frac{1}{2+a}}^{\frac{1}{2-a}}\quad(\because\text{(式3)}) \\&=\pi\left\{\left(\dfrac{1}{2-a}+\dfrac{1}{2+a}\right)-\dfrac{1}{3}\left(\left(\dfrac{1}{2-a}\right)^3+\left(\dfrac{1}{2+a}\right)^3\right)\right\} \end{align}

である．

が求める体積である．