3重積分重心密度、一様、物体、楕円、上半分、三角関数(正弦と余弦)、累乗、置換積分法

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9章(3重積分)、3(重心)の練習問題3の解答を求めてみる。

\begin{array}{l} a, b > 0 \\ - a \leq x \leq a \\ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ y^{2} = (1 - \frac{x^{2}}{a^{2}}) b^{2} \\ = \frac{b^{2}}{a^{2}} (a^{2} - x^{2}) \\ y = \frac{b}{a} \sqrt{a^{2} - x^{2}} \end{array}

よって、

\begin{array}{l} m \\ = 2 \frac{b}{a} \int_{0}^{a} \sqrt{a^{2} - x^{2}} dx \\ x = a (\sin t) \\ 0 \leq t \leq \frac{π}{2} \\ \frac{dx}{dt} = a (\cos t) \\ dx = a (\cos t) dt \\ m \\ = \frac{2 b}{a} \int_{0}^{\frac{π}{2}} \sqrt{a^{2} - a^{2} \sin^{2} t} a (\cos t) dt \\ = \frac{2 b}{a} \int_{0}^{\frac{π}{2}} a \sqrt{1 - \sin^{2} t} a (\cos t) dt \\ = 2 a b \int_{0}^{\frac{π}{2}} \sqrt{\cos^{2} t} \cos t dt \\ = 2 a b \int_{0}^{\frac{π}{2}} \cos^{2} t dt \\ = 2 a b \cdot \frac{π}{2} \cdot \frac{1}{2} \\ = \frac{π a b}{2} \end{array}

重心の各座標について。

\begin{array}{l} \bar{x} \\ = \frac{1}{m} \int_{- a}^{a} \int_{0}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} x dy dx \\ = \frac{2}{π a b} \int_{- a}^{a} x \cdot \frac{b}{a} \sqrt{a^{2} - x^{2}} dx \\ = \frac{2}{π a^{2}} \int_{- \frac{π}{2}}^{\frac{π}{2}} a (\sin t) \cos^{2} t dt \\ = \frac{2}{π a} \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin t (1 - \sin^{2} t) dt \\ = \frac{2}{π a} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\sin t - \sin^{3} t) dt \\ = 0 \end{array}

また、

\begin{array}{l} \bar{y} \\ = \frac{1}{m} \int_{- a}^{a} \int_{0}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} y dy dx \\ = \frac{1}{π a b} \int_{- a}^{a} {(\frac{b}{a} \sqrt{a^{2} - x^{2}})}^{2} dx \\ = \frac{2}{π a b} \cdot \frac{b^{2}}{a^{2}} \int_{0}^{a} (a^{2} - x^{2}) dx \\ = \frac{2 b}{π a^{3}} {[a^{2} x - \frac{1}{3} x^{3}]}_{0}^{a} \\ = \frac{2 b}{π a^{3}} (a^{3} - \frac{1}{3} a^{3}) \\ = \frac{2 b}{π a^{3}} \cdot \frac{2 a^{3}}{3} \\ = \frac{4 b}{3 π} \end{array}