【解答】
図 1: 楕円と接線 ,および 軸周りに回転させる領域(斜線部)
対称性から,立体のうち の部分の体積 ,もとめる体積 として,
である.接線は 軸平行でないので,傾きをとして
とおける.これが楕円と接するので, とおき直した座標で,
と が接する.したがって円の中心と直線の距離がだから
各辺以上なので2乗して,
となる.
図から接線の傾きは負だから, より複号負をとって,
である.
この時接点は となる.
図のように各点をおくと,は台形を軸周りに一回転してできる円錐の一部から,楕円弧を軸周りに回転してできる立体の体積を引いたものである.
数式で表すと
\begin is only supported in math mode\begin{align} V_1 &= \left(\text{ \begin{tikzpicture}[baseline={(0,0.3)}, scale=0.85, >=stealth] \fill[pattern=north east lines, pattern color=black!50] (0,0) -- (1.2,0) -- (0.3,0.7) -- (0,0.7) -- cycle; \draw[thick] (0,0) -- (1.2,0) -- (0.3,0.7) -- (0,0.7) -- cycle; \draw[dashed, gray!70] (0,0) -- (0,0.9); \draw[->, thick, red!80!black] (-0.2, 1.0) arc[start angle=150, end angle=390, x radius=0.2, y radius=0.06]; \node[below left, font=\scriptsize, inner sep=1pt] at (0,0) {$O$}; \node[below right, font=\scriptsize, inner sep=1pt] at (1.2,0) {$A$}; \node[above right, font=\scriptsize, inner sep=1pt] at (0.3,0.7) {$P$}; \node[left, font=\scriptsize, inner sep=1pt] at (0,0.7) {$R$}; \node[circle, fill, inner sep=0.6pt] at (0,0) {}; \node[circle, fill, inner sep=0.6pt] at (1.2,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.3,0.7) {}; \node[circle, fill, inner sep=0.6pt] at (0,0.7) {}; \end{tikzpicture}}\right) - \left(\text{ \begin{tikzpicture}[baseline={(0,0.3)}, scale=0.85, >=stealth] \fill[pattern=north east lines, pattern color=black!50] (0,0) -- (0.8,0) arc[start angle=0, end angle=60, x radius=0.8, y radius=0.8] -- (0,0.7) -- cycle; \draw[thick] (0,0) -- (0.8,0) arc[start angle=0, end angle=60, x radius=0.8, y radius=0.8] -- (0,0.7) -- cycle; \draw[dashed, gray!70] (0,0) -- (0,0.9); \draw[->, thick, red!80!black] (-0.2, 1.0) arc[start angle=150, end angle=390, x radius=0.2, y radius=0.06]; \node[below left, font=\scriptsize, inner sep=1pt] at (0,0) {$O$}; \node[below right, font=\scriptsize, inner sep=1pt] at (0.8,0) {$Q$}; \node[above right, font=\scriptsize, inner sep=1pt] at (0.4,0.7) {$P$}; \node[left, font=\scriptsize, inner sep=1pt] at (0,0.7) {$R$}; \node[circle, fill, inner sep=0.6pt] at (0,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.8,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.4,0.7) {}; \node[circle, fill, inner sep=0.6pt] at (0,0.7) {}; \end{tikzpicture}}\right) \end{align}
である.
各項を個別に計算しよう.まず第一項は,を一回転してできる円錐からを一回転してできる円錐を差し引いたものである.
これらの辺の長さは下図の通りである.
図 2: 円錐部分の体積計算に用いる直角三角形の相似
従ってこの体積は
\begin is only supported in math mode\begin{align} \begin{split} \left( \text{ \begin{tikzpicture}[baseline={(0,0.3)}, scale=0.85, >=stealth] \fill[pattern=north east lines, pattern color=black!50] (0,0) -- (1.2,0) -- (0.3,0.7) -- (0,0.7) -- cycle; \draw[thick] (0,0) -- (1.2,0) -- (0.3,0.7) -- (0,0.7) -- cycle; \draw[dashed, gray!70] (0,0) -- (0,0.9); \draw[->, thick, red!80!black] (-0.2, 1.0) arc[start angle=150, end angle=390, x radius=0.2, y radius=0.06]; \node[below left, font=\scriptsize, inner sep=1pt] at (0,0) {$O$}; \node[below right, font=\scriptsize, inner sep=1pt] at (1.2,0) {$A$}; \node[above right, font=\scriptsize, inner sep=1pt] at (0.3,0.7) {$P$}; \node[left, font=\scriptsize, inner sep=1pt] at (0,0.7) {$R$}; \node[circle, fill, inner sep=0.6pt] at (0,0) {}; \node[circle, fill, inner sep=0.6pt] at (1.2,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.3,0.7) {}; \node[circle, fill, inner sep=0.6pt] at (0,0.7) {}; \end{tikzpicture}} \right) &= \dfrac{1}{3}\pi(2a)^2\dfrac{2\sqrt{3}}{3}b-\dfrac{1}{3}\pi\left(\dfrac{1}{2}a\right)^2\left(\dfrac{2\sqrt{3}}{3}b-\dfrac{\sqrt{3}}{2}b\right) \\ &= \dfrac{1}{3}\pi\left[\dfrac{8\sqrt{3}}{3}a^2b-\dfrac{\sqrt{3}}{24}a^2b\right] \\ &= \dfrac{7\sqrt{3}}{8}\pi a^2b \end{split} \end{align}
である.
次に(式7)の第二項の楕円弧を回転した部分の体積は
\begin is only supported in math mode\begin{align} \begin{split} \left( \text{ \begin{tikzpicture}[baseline={(0,0.3)}, scale=0.85, >=stealth] \fill[pattern=north east lines, pattern color=black!50] (0,0) -- (0.8,0) arc[start angle=0, end angle=60, x radius=0.8, y radius=0.8] -- (0,0.7) -- cycle; \draw[thick] (0,0) -- (0.8,0) arc[start angle=0, end angle=60, x radius=0.8, y radius=0.8] -- (0,0.7) -- cycle; \draw[dashed, gray!70] (0,0) -- (0,0.9); \draw[->, thick, red!80!black] (-0.2, 1.0) arc[start angle=150, end angle=390, x radius=0.2, y radius=0.06]; \node[below left, font=\scriptsize, inner sep=1pt] at (0,0) {$O$}; \node[below right, font=\scriptsize, inner sep=1pt] at (0.8,0) {$Q$}; \node[above right, font=\scriptsize, inner sep=1pt] at (0.4,0.7) {$P$}; \node[left, font=\scriptsize, inner sep=1pt] at (0,0.7) {$R$}; \node[circle, fill, inner sep=0.6pt] at (0,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.8,0) {}; \node[circle, fill, inner sep=0.6pt] at (0.4,0.7) {}; \node[circle, fill, inner sep=0.6pt] at (0,0.7) {}; \end{tikzpicture}} \right) &= \pi\int_0^{\frac{\sqrt{3}}{2}b}a^2\left(1-\dfrac{y^2}{b^2}\right)dy \\ &=a^2\pi\left[y-\dfrac{1}{3b^2}y^3\right]_0^{\frac{\sqrt{3}}{2}b} \\ &=\dfrac{3\sqrt{3}}{8}\pi a^2b \end{split} \end{align}
である.
(式8),(式9)を(式7)に代入して
だから,(式1)に代入して求める体積は
である.