グリーンの定理(Green's theorem)
閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。
$$
\begin{eqnarray}
\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
&=&
\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\end{eqnarray}
$$
面積分を周回積分へ変形し，区間毎の積分に展開する
Dの面積Aは面積分で表すと
$$
\begin{eqnarray}
A&=&\iint_D 1\;\mathrm{d}x\mathrm{d}y
\end{eqnarray}
$$
であり，これをグリーンの定理の式で満たすために例えば
$$
\begin{eqnarray}
Q(x, y)&=&x
\\P(x, y)&=&0
\end{eqnarray}
$$
とおく(それぞれ一階微分可能で連続(\(C^1\)級凾数))．これは
$$
\begin{eqnarray}
\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&1-0
\\&=&1
\end{eqnarray}
$$
となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この\(P,Q\)を用いて周回積分側の 被積分凾数を求めると以下のようになる．
$$
\begin{eqnarray}
A&=&\iint_D 1\;\mathrm{d}x\mathrm{d}y
\\&=&\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
\\&=&\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\\&=&\oint_C 0\mathrm{d}x+x\mathrm{d}y
\\&=&\oint_C x\mathrm{d}y
\end{eqnarray}
$$
ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする．
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1),(x_2, y_2), \cdots, (x_n, y_n)=(x_0, y_0)
\end{eqnarray}
$$
周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる．
$$
\begin{eqnarray}
A&=&\oint_C x\mathrm{d}y
\\&=&\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} x \mathrm{d}y
\end{eqnarray}
$$
多角形の頂点間線分(辺)に重なる直線の式を求める
多角形の各頂点において今の頂点\((x_k, y_k)\)とその前の頂点\((x_{k-1}, y_{k-1})\)とを通る直線の式は以下のように求められる．
$$
\begin{eqnarray}
y_-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)
\;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p)
\\\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)&=&x-x_{k-1}
\\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1}
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}\color{red}{+x_{k-1}y_{k-1}} \color{black}{+x_{k-1}y_k} \color{red}{-x_{k-1}y_{k-1}} }{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}}
\end{eqnarray}
$$
区間毎の積分より面積を求める
区間毎の積分の和の式における被積分凾数に，先ほど求めた直線の式を代入する．
$$
\begin{eqnarray}
A&=&\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} x \mathrm{d}y
\\&=&\sum_{k=1}^n
\int^{y_k}_{y_{k-1}} \left( \frac{x_k - x_{k-1}}{y_k - y_{k-1}}y + \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right) \mathrm{d}y
\;\cdots\;x=\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}}
\\&=&\sum_{k=1}^n\left\{
\int^{y_k}_{y_{k-1}} \frac{x_k - x_{k-1}}{y_k - y_{k-1}}y \mathrm{d}y
+ \int^{y_k}_{y_{k-1}} \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \mathrm{d}y
\right\}
\;\cdots\;\int_X(A(x)+B(x))\mathrm{d}x=\int_XA(x)\mathrm{d}x+\int_XB(x)\mathrm{d}x
\\&=&\sum_{k=1}^n\left\{
\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\int^{y_k}_{y_{k-1}} y \mathrm{d}y
+ \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \int^{y_k}_{y_{k-1}} \mathrm{d}y
\right\}
\;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x
\\&=&\sum_{k=1}^n\left\{
\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\left[\frac{1}{2}y^2\right]^{y_k}_{y_{k-1}}
+ \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \left[y\right]^{y_k}_{y_{k-1}}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2}\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\left[y^2\right]^{y_k}_{y_{k-1}}
+ \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \left[y\right]^{y_k}_{y_{k-1}}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2}\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\left(y_k^2-y_{k-1}^2\right)
+ \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \left(y_k-y_{k-1}\right)
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2}\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\left(y_k+y_{k-1}\right)\left(y_k-y_{k-1}\right)
+ \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \left(y_k-y_{k-1}\right)
\right\}
\;\cdots\;A^2-B^2=(A+B)(A-B)
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2} (x_k - x_{k-1}) \left( y_k+y_{k-1} \right)
+ x_{k-1}y_k - x_ky_{k-1}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2}\left( x_ky_k+x_ky_{k-1}-x_{k-1}y_k-x_{k-1}y_{k-1} \right)
+ x_{k-1}y_k - x_ky_{k-1}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2} x_ky_k
+\left(\frac{1}{2}-1\right)x_ky_{k-1}
+\left(-\frac{1}{2}+1\right)x_{k-1}y_k
-\frac{1}{2} x_{k-1}y_{k-1}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2} x_ky_k
-\frac{1}{2}x_ky_{k-1}
+\frac{1}{2}x_{k-1}y_k
-\frac{1}{2} x_{k-1}y_{k-1}
\right\}
\\&=&\sum_{k=1}^n\left\{
\frac{1}{2}\left( x_ky_k-x_ky_{k-1}+x_{k-1}y_k-x_{k-1}y_{k-1} \right)
\right\}
\\&=&\sum_{k=1}^n\left[
\frac{1}{2}\left\{ x_k\left(y_k-y_{k-1}\right)+x_{k-1}\left(y_k-y_{k-1}\right) \right\}
\right]
\\&=&\sum_{k=1}^n
\frac{1}{2}\left(x_k+x_{k-1}\right)\left(y_k-y_{k-1}\right)
\\&=&\frac{1}{2}\sum_{k=1}^n
\left(x_k+x_{k-1}\right)\left(y_k-y_{k-1}\right)
\end{eqnarray}
$$
多角形の頂点座標から，その面積が求められた．
例:正四角形
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1),(x_2, y_2), (x_3, y_3),(x_4=x_0, y_4=y_0)&=&
(0, 0),(1, 0), (1, 1), (0, 1), (0,0)
\end{eqnarray}
$$
$$
\begin{eqnarray}
A&=&\frac{1}{2}\sum_{k=1}^n
\left(x_k+x_{k-1}\right)\left(y_k-y_{k-1}\right)
\\&=&\frac{1}{2}\left\{
\left(x_1+x_0\right)\left(y_1-y_0\right)
+\left(x_2+x_1\right)\left(y_2-y_1\right)
+\left(x_3+x_2\right)\left(y_3-y_2\right)
+\left(x_4+x_3\right)\left(y_4-y_3\right)
\right\}
\\&=&\frac{1}{2}\left\{
\left(1+0\right)\left(0-0\right)
+\left(1+1\right)\left(1-0\right)
+\left(0+1\right)\left(1-1\right)
+\left(0+0\right)\left(0-1\right)
\right\}
\\&=&\frac{1}{2}\left(
0+2+0+0
\right)
\\&=&\frac{1}{2}2
\\&=&1
\end{eqnarray}
$$
例:二等辺三角形
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1),(x_2, y_2),(x_3=x_0, y_3=y_0)&=&
(0, 0),(1, 0), (\frac{1}{2}, 1),(0, 0)
\end{eqnarray}
$$
$$
\begin{eqnarray}
A&=&\frac{1}{2}\sum_{k=1}^n
\left(x_k+x_{k-1}\right)\left(y_k-y_{k-1}\right)
\\&=&\frac{1}{2}\left\{
\left(x_1+x_0\right)\left(y_1-y_0\right)
+\left(x_2+x_1\right)\left(y_2-y_1\right)
+\left(x_3+x_2\right)\left(y_3-y_2\right)
\right\}
\\&=&\frac{1}{2}\left\{
\left(1+0\right)\left(0-0\right)
+\left(\frac{1}{2}+1\right)\left(1-0\right)
+\left(0+\frac{1}{2}\right)\left(0-1\right)
\right\}
\\&=&\frac{1}{2}\left(
0+\frac{3}{2}-\frac{1}{2}
\right)
\\&=&\frac{1}{2}1
\\&=&\frac{1}{2}
\end{eqnarray}
$$
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