2重積分反復積分積、累乗、三角関数(正弦と余弦と正接と正割)

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)、2(反復積分)の練習問題1の解答を求めてみる。

\begin{array}{l} \int_{0}^{2} \int_{1}^{3} (x + y) dx dy \\ = \int_{0}^{2} {[\frac{1}{2} x^{2} + y x]}_{1}^{3} dy \\ = \int_{0}^{2} (4 + 2 y) dy \\ = {[4 y + y^{2}]}_{0}^{2} \\ = 12 \end{array}

\begin{array}{l} \int_{0}^{2} \int_{1}^{x^{2}} y dy dx \\ = \int_{0}^{2} {[\frac{1}{2} y^{2}]}_{1}^{x^{2}} dx \\ = \frac{1}{2} \int_{0}^{2} (x^{4} - 1) dx \\ = \frac{1}{2} {[\frac{1}{5} x^{5} - x]}_{0}^{2} \\ = \frac{1}{2} (\frac{32}{5} - 2) \\ = \frac{16}{5} \\ = \frac{11}{5} \end{array}

\begin{array}{l} \int_{0}^{1} \int_{y^{2}}^{y} \sqrt{x} dx dy \\ = \frac{2}{3} \int_{0}^{1} {[x^{\frac{3}{2}}]}_{y^{2}}^{y} dy \\ = \frac{2}{3} \int_{0}^{1} (y^{\frac{3}{2}} - y^{3}) dy \\ = \frac{2}{3} {[\frac{2}{5} y^{\frac{5}{2}} - \frac{1}{4} y^{4}]}_{0}^{1} \\ = \frac{2}{3} (\frac{2}{5} - \frac{1}{4}) \\ = \frac{2}{3} \cdot \frac{8 - 5}{20} \\ = \frac{1}{10} \end{array}

\begin{array}{l} \int_{0}^{π} \int_{0}^{x} x \sin y dy dx \\ = \int_{0}^{π} x {[- \cos y]}_{0}^{x} d x \\ = \int_{0}^{π} x (- \cos x + 1) dx \\ = - \int_{0}^{π} x \cos x d x + {[\frac{1}{2} x^{2}]}_{0}^{π} \\ = {[- x \sin x]}_{0}^{π} + \int_{0}^{π} \sin x dx + \frac{1}{2} π^{2} \\ = {[- \cos x]}_{0}^{π} + \frac{1}{2} π^{2} \\ = 2 + \frac{1}{2} π^{2} \end{array}

\begin{array}{l} \int_{1}^{2} \int_{y}^{y^{2}} dx dy \\ = \int_{1}^{2} (y^{2} - y) dy \\ = {[\frac{1}{3} y^{3} - \frac{1}{2} y^{2}]}_{1}^{2} \\ = \frac{7}{3} - \frac{3}{2} \\ = \frac{14 - 9}{6} \\ = \frac{5}{6} \end{array}

\begin{array}{l} \int_{0}^{π} \int_{0}^{\sin x} y dy dx \\ = \frac{1}{2} \int_{0}^{π} {[y^{2}]}_{0}^{\sin x} d x \\ = \frac{1}{2} \int_{0}^{π} \sin^{2} x dx \\ = \frac{1}{2} \int_{0}^{π} \frac{1 - \cos 2 x}{2} dx \\ = \frac{1}{4} {[x - \frac{1}{2} \sin 2 x]}_{0}^{π} \\ = \frac{π}{4} \end{array}

\begin{array}{l} \int_{0}^{\frac{π}{2}} \int_{0}^{2} r^{2} \cos θ d r d θ \\ = \frac{1}{3} \int_{0}^{\frac{π}{2}} {[r^{3}]}_{0}^{2} \cos θ d θ \\ = \frac{8}{3} {[\sin θ]}_{0}^{\frac{π}{2}} \\ = \frac{8}{3} \end{array}

\begin{array}{l} \int_{0}^{2 π} \int_{0}^{1 - \cos θ} r^{3} \cos^{2} θ d r d θ \\ = \frac{1}{4} \int_{0}^{2 π} {[r^{4}]}_{0}^{1 - \cos θ} \cos^{2} θ d θ \\ = \frac{1}{4} \int_{0}^{2 π} {(1 - \cos θ)}^{4} \cos^{2} θ d θ \\ = \frac{1}{4} \int_{0}^{2 π} (\cos^{6} θ - 4 \cos^{5} θ + 6 \cos^{4} θ - 4 \cos^{3} θ + \cos^{2} θ) d θ \\ = \int_{0}^{\frac{π}{2}} (\cos^{6} θ + 6 \cos^{4} θ + \cos^{2} θ) d θ \\ = \frac{π}{2} (\frac{5 \cdot 3}{6 \cdot 4 \cdot 2} + 6 \cdot \frac{3}{4 \cdot 2} + \frac{1}{2}) \\ = \frac{π}{2} (\frac{5}{16} + \frac{9}{4} + \frac{1}{2}) \\ = \frac{(5 + 36 + 8) π}{32} \\ = \frac{49 π}{32} \end{array}

\begin{array}{l} \int_{0}^{\arctan \frac{3}{2}} \int_{0}^{2 \sec θ} r d r d θ \\ = \frac{1}{2} \int_{0}^{\arctan \frac{3}{2}} {(2 \sec θ)}^{2} d θ \\ = 2 \int_{0}^{\arctan \frac{3}{2}} \sec^{2} θ d θ \\ = 2 {[\tan θ]}_{0}^{\arctan \frac{3}{2}} \\ = 2 \cdot \frac{3}{2} \\ = 3 \end{array}