3重積分重心密度、一様、物体、2つの放物線に囲まれる領域

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4の解答を求めてみる。

\begin{array}{l} 2 x - x^{2} = 3 x^{2} - 6 x \\ 4 x^{2} - 8 x = 0 \\ x (x - 2) = 0 \end{array}

よって、

\begin{array}{l} m = \int_{0}^{2} ((2 x - x^{2}) - (3 x^{2} - 6 x)) dx \\ = \int_{0}^{2} (- 4 x^{2} + 8 x) dx \\ = - 4 \int_{0}^{2} (x^{2} - 2 x) dx \\ = - 4 {[\frac{1}{3} x^{3} - x^{2}]}_{0}^{2} \\ = - 4 (\frac{8}{3} - 4) \\ = - 4 \cdot \frac{- 4}{3} \\ = \frac{16}{3} \end{array}

重心の各座標。

\begin{array}{l} \bar{x} \\ = \frac{1}{m} \int_{0}^{2} \int_{3 x^{2} - 6 x}^{2 x - x^{2}} x dy dx \\ = \frac{3}{16} \int_{0}^{2} x ((2 x - x^{2}) - (3 x^{2} - 6 x)) dx \\ = \frac{3}{16} \int_{0}^{2} x (4 x^{2} - 8 x) dx \\ = \frac{3}{4} \int_{0}^{2} (x^{3} - 2 x^{2}) dx \\ = \frac{3}{4} {[\frac{1}{4} x^{4} - \frac{2}{3} x^{3}]}_{0}^{2} \\ = \frac{3}{4} \cdot 2^{4} (\frac{1}{4} - \frac{1}{3}) \\ = 3 \cdot 2^{2} \cdot \frac{1}{12} \\ = 1 \end{array}

また、

\begin{array}{l} \bar{y} \\ = \frac{1}{m} \int_{0}^{2} \int_{3 x^{2} - 6 x}^{2 x - x^{2}} y dy dx \\ = \frac{3}{16} \cdot \frac{1}{2} \int_{0}^{2} ({(2 x - x^{2})}^{2} - {(3 x^{2} - 6 x)}^{2}) dx \\ = \frac{3}{2^{5}} \int_{0}^{2} (4 x^{2} - 4 x^{3} + x^{4} - 9 x^{4} + 36 x^{3} - 36 x^{2}) dx \\ = \frac{3}{2^{5}} \int_{0}^{2} (- 8 x^{4} + 32 x^{3} - 32 x^{2}) dx \\ = \frac{3}{2^{2}} \int_{0}^{2} (- x^{4} + 4 x^{3} - 4 x^{2}) dx \\ = \frac{3}{2^{2}} {[- \frac{1}{5} x^{5} + x^{4} - \frac{4}{3} x^{3}]}_{0}^{2} \\ = \frac{3}{2^{2}} (- \frac{1}{5} \cdot 2^{5} + 2^{4} - \frac{2^{5}}{3}) \\ = 2^{2} \cdot 3 (- \frac{2}{5} + 1 - \frac{2}{3}) \\ = 2^{2} \cdot 3 \cdot \frac{- 6 + 15 - 10}{15} \\ = - \frac{4}{5} \end{array}