2重積分極座標領域、三角関数(正弦と余弦)、倍角、立体、体積

学習環境

続解析入門 (原書第2版) (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第7章(2重積分)、3(極座標)の練習問題14の解答を求めてみる。

\begin{array}{l} 2 \int_{- \frac{π}{4}}^{\frac{π}{4}} \int_{0}^{\sqrt{2} a \sqrt{\cos 2 θ}} \sqrt{2 a^{2} - r^{2}} r d r d θ \\ = 2 \cdot \frac{2}{3} (- \frac{1}{2}) \int_{- \frac{π}{4}}^{\frac{π}{4}} {[{(2 a^{2} - r^{2})}^{\frac{3}{2}}]}_{0}^{a \sqrt{2 \cos 2 θ}} d θ \\ = - \frac{2}{3} \int_{- \frac{π}{4}}^{\frac{π}{4}} ({(2 a^{2} - 2 a^{2} \cos 2 θ)}^{\frac{3}{2}} - {(2 a^{2})}^{\frac{3}{2}}) d θ \\ = - \frac{4 \sqrt{2}}{3} a^{3} (\int_{- \frac{π}{4}}^{\frac{π}{4}} {(1 - \cos 2 θ)}^{\frac{3}{2}} d θ - {[θ]}_{- \frac{π}{4}}^{\frac{π}{4}}) \\ = - \frac{4 \sqrt{2}}{3} a^{3} (\int_{- \frac{π}{4}}^{\frac{π}{4}} {(1 - \cos 2 θ)}^{\frac{3}{2}} d θ - \frac{π}{2}) \\ = - \frac{4 \sqrt{2}}{3} a^{3} (\int_{- \frac{π}{4}}^{\frac{π}{4}} {(1 - (\cos^{2} θ - \sin^{2} θ))}^{\frac{3}{2}} d θ - \frac{π}{2}) \\ = \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} - \int_{- \frac{π}{4}}^{\frac{π}{4}} {(1 - (1 - 2 \sin^{2} θ))}^{\frac{3}{2}} d θ) \\ = \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} - \int_{- \frac{π}{4}}^{\frac{π}{4}} {(2 \sin^{2} θ)}^{\frac{3}{2}} d θ) \\ = \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} - 4 \sqrt{2} \int_{0}^{\frac{π}{4}} \sin^{3} θ d θ) \end{array}

\begin{array}{l} \int \sin^{3} θ d θ \\ = - \cos θ \sin^{2} θ + 2 \int \cos^{2} θ \sin θ d θ \\ = - \sin^{2} θ \cos θ + 2 \int (1 - \sin^{2} θ) \sin θ d θ \\ = - \sin^{2} θ \cos θ - 2 \cos θ - 2 \int \sin^{3} θ d θ \\ \int \sin^{3} θ d θ = \frac{1}{3} (- \sin^{2} θ \cos θ - 2 \cos θ) \end{array}

\begin{array}{l} \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} + \frac{4 \sqrt{2}}{3} {[\sin^{2} θ \cos θ + 2 \cos θ]}_{0}^{\frac{π}{4}}) \\ = \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} + \frac{4 \sqrt{2}}{3} (\frac{1}{2 \sqrt{2}} + \frac{2}{\sqrt{2}} - 2)) \\ = \frac{4 \sqrt{2} a^{3}}{3} (\frac{π}{2} + \frac{2}{3} + \frac{8}{3} - \frac{8 \sqrt{2}}{3}) \\ = \frac{4 \sqrt{2} a^{3}}{9} (\frac{3}{2} π + 10 - 8 \sqrt{2}) \\ = \frac{2 \sqrt{2} a^{3}}{9} (3 π + 20 - 16 \sqrt{2}) \end{array}

コード、入出力結果(Wolfram Language, Jupyter Notebook)

a = 2

SphericalPlot3D[
    Sqrt[2a^2-r^2] r,
    {θ, 0, 2Pi},
    {r, 0, Sqrt[2a^2 Cos[2θ]]}
]