3重積分重心密度、一様、物体、極座標、三角関数、正弦と余弦、累乗、曲線、板状領域、原点からの距離、比例、質量

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9の解答を求めてみる。

質量。

\begin{array}{l} m \\ = \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} (k r) r d r d θ \\ = k \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} r^{2} d r d θ \\ = \frac{k}{3} \int_{0}^{π} {(2 (1 + \cos θ))}^{3} d θ \\ = \frac{8 k}{3} \int_{0}^{π} (1 + 3 \cos θ + 3 \cos^{2} θ + \cos^{3} θ) d θ \\ = \frac{8 k}{3} (π + 3 \int_{0}^{π} \cos^{2} θ d θ) \\ = \frac{8 k}{3} (π + 6 \int_{0}^{\frac{π}{2}} \cos^{2} θ d θ) \\ = \frac{8 k}{3} (π + 6 \cdot \frac{π}{2} \cdot \frac{1}{2}) \\ = \frac{8 k}{3} (π + \frac{3 π}{2}) \\ = \frac{8 k}{3} \cdot \frac{5 π}{2} \\ = \frac{20 k π}{3} \end{array}

重心。

各座標。

\begin{array}{l} \bar{x} \\ = \frac{1}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} x (k r) r d r d θ \\ = \frac{k}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} (r \cos θ) r^{2} d r d θ \\ = \frac{k}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} r^{3} \cos θ d r d θ \\ = \frac{k}{4 m} \int_{0}^{π} {(2 (1 + \cos θ))}^{4} \cos θ d θ \\ = \frac{4 k}{m} \int_{0}^{π} (\cos θ + 4 \cos^{2} θ + 6 \cos^{3} θ + 4 \cos^{4} θ + \cos^{5} θ) d θ \\ = \frac{4 k}{m} \int_{0}^{π} (4 \cos^{2} θ + 4 \cos^{4} θ) d θ \\ = \frac{4 \cdot 4 k}{m} \cdot 2 \int_{0}^{\frac{π}{2}} (\cos^{2} θ + \cos^{4} θ) d θ \\ = \frac{3 \cdot 4 \cdot 4 \cdot 2 k}{20 k π} (\frac{π}{2} \cdot \frac{1}{2} + \frac{π}{2} \cdot \frac{3}{4 \cdot 2}) \\ = \frac{3 \cdot 4 \cdot 2}{20} (1 + \frac{3}{4}) \\ = \frac{3 \cdot 4 \cdot 2}{20} \cdot \frac{7}{4} \\ = \frac{21}{10} \end{array}

また、

\begin{array}{l} \bar{y} \\ = \frac{1}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} y (k r) r d r d θ \\ = \frac{k}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} (r \sin θ) r^{2} d r d θ \\ = \frac{k}{m} \int_{0}^{π} \int_{0}^{2 (1 + \cos θ)} r^{3} \sin θ d r d θ \\ = \frac{k}{4 m} \int_{0}^{π} {(2 (1 + \cos θ))}^{4} \sin θ d θ \\ = \frac{4 k}{m} \int_{0}^{π} (1 + 4 \cos θ + 6 \cos^{2} θ + 4 \cos^{3} θ + \cos^{4} θ) \sin θ d θ \\ = \frac{4 k}{m} ( \\ + \int_{0}^{π} \sin θ d θ \\ + 2 \int_{0}^{π} \sin 2 θ d θ \\ + 6 \int_{0}^{π} (\sin θ - \sin^{3} θ) d θ \\ + 4 \int_{0}^{π} \sin θ \cos^{3} θ d θ \\ + \int_{0}^{π} {(1 - \sin^{2} θ)}^{2} \sin θ d θ) \\ = \frac{4 k}{m} ( \\ + {[- \cos θ]}_{0}^{π} \\ + {[- \cos 2 θ]}_{0}^{π} \\ + 6 ({[- \cos θ]}_{0}^{π} - 2 \int_{0}^{\frac{π}{2}} \sin^{3} θ d θ) \\ + 4 \int_{0}^{π} \sin θ \cos^{3} θ d θ \\ + \int_{0}^{π} (\sin θ - 2 \sin^{3} θ + \sin^{5} θ) d θ) \\ = \frac{4 k}{m} (2 + 6 (2 - 2 \cdot \frac{2}{3}) + 2 (1 - 2 \cdot \frac{2}{3} + \frac{4 \cdot 2}{5 \cdot 3}) + 4 \int_{0}^{π} \sin θ \cos^{3} θ d θ) \\ = \frac{4 k}{m} (6 + 2 (1 - \frac{4}{3} + \frac{8}{15}) + 4 \int_{0}^{π} \sin θ \cos^{3} θ d θ) \end{array}

分けて計算。

\begin{array}{l} \int \sin θ \cos^{3} θ d θ \\ = - \cos^{4} θ + \int \cos θ (3 \cos^{2} θ) (- \sin θ) d θ \\ = - \cos^{4} θ - 3 \int \sin θ \cos^{3} θ d θ \\ \int \sin θ \cos^{3} θ d θ = - \frac{3}{4} \cos^{4} θ \end{array}

よって、

\begin{array}{l} \bar{y} \\ = \frac{4 k}{m} (6 + 2 \cdot \frac{15 - 20 + 8}{15} - 3 {[\cos^{4} θ]}_{0}^{π}) \\ = \frac{4 k}{m} (6 + \frac{2 \cdot 3}{15}) \\ = \frac{4 k}{m} (6 + \frac{2}{5}) \\ = \frac{3}{20 k π} \cdot 4 k \cdot \frac{32}{5} \\ = \frac{96}{25 π} \end{array}