グリーンの定理グリーンの定理の標準形正方形の周、円、半径、楕円

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8章(グリーンの定理)、1(グリーンの定理の標準形)の練習問題1の解答を求めてみる。

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \iint_{R} (1 - 2 y) dx dy \\ = \int_{0}^{2} \int_{0}^{2} (1 - 2 y) dx dy \\ = \int_{0}^{2} 2 (1 - 2 y) dy \\ = {[2 y - 2 y^{2}]}_{0}^{2} \\ = 4 - 8 \\ = - 4 \end{array}

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \int_{- 1}^{1} \int_{- 1}^{1} (1 - 2 y) dx dy \\ = \int_{- 1}^{1} (1 - 2 y) 2 dy \\ = {[2 y - 2 y^{2}]}_{- 1}^{1} \\ = 4 \end{array}

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \iint_{x^{2} + y^{2} = 4} (1 - 2 y) dx dy \\ = 2^{2} π \\ = 4 π \end{array}

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \iint_{x^{2} + y^{2} = 1} (1 - 2 y) dx dy \\ = π \end{array}

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \int_{- 2}^{2} \int_{- 2}^{2} (1 - 2 y) dy dx \\ = \int_{- 2}^{2} {[y - y^{2}]}_{- 2}^{2} dx \\ = 4 \int_{- 2}^{2} dy \\ = 4 \cdot 4 \\ = 16 \end{array}

\begin{array}{l} \int_{C} y^{2} dx + x dy \\ = \iint_{\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1} (1 - 2 y) dx dy \\ = π a b \end{array}