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の重心のX座標\(X_G\)は
$$
\begin{eqnarray}
X_G&=&\iint_D \rho(x,y)x\;\mathrm{d}x\mathrm{d}y
\\&=&\iint_D \frac{1}{A}x\;\mathrm{d}x\mathrm{d}y
\;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積)
\\&=&\frac{1}{A}\iint_D x\;\mathrm{d}x\mathrm{d}y
\;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x
\end{eqnarray}
$$
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
$$
\begin{eqnarray}
Q(x, y)&=&\frac{1}{2}x^2
\\P(x, y)&=&0
\end{eqnarray}
$$
とおく(それぞれ一階微分可能で連続(\(C^1\)級凾数))．これは
$$
\begin{eqnarray}
\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&x-0
\\&=&x
\end{eqnarray}
$$
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．
$$
\begin{eqnarray}
X_G&=&\frac{1}{A}\iint_D x\;\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\\&=&\frac{1}{A}\oint_C 0\mathrm{d}x+\frac{1}{2}x^2\mathrm{d}y
\\&=&\frac{1}{A}\oint_C \frac{1}{2}x^2\mathrm{d}y
\\&=&\frac{1}{2A}\oint_C x^2\mathrm{d}y
\end{eqnarray}
$$
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．
$$
\begin{eqnarray}
X_G&=&\frac{1}{2A}\oint_C x^2\mathrm{d}y
\\&=&\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} x^2 \mathrm{d}y
\end{eqnarray}
$$
同様にDの重心のY座標\(Y_G\)は
$$
\begin{eqnarray}
Y_G&=&\iint_D \rho(x,y)y\;\mathrm{d}x\mathrm{d}y
\\&=&\iint_D \frac{1}{A}y\;\mathrm{d}x\mathrm{d}y
\;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積)
\\&=&\frac{1}{A}\iint_D y\;\mathrm{d}x\mathrm{d}y
\;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x
\end{eqnarray}
$$
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
$$
\begin{eqnarray}
Q(x, y)&=&0
\\P(x, y)&=&-\frac{1}{2}y^2
\end{eqnarray}
$$
とおく(それぞれ一階微分可能で連続(\(C^1\)級凾数))．これは
$$
\begin{eqnarray}
\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&0-(-y)
\\&=&y
\end{eqnarray}
$$
となり，やは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．
$$
\begin{eqnarray}
Y_G&=&\frac{1}{A}\iint_D y\;\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\\&=&\frac{1}{A}\oint_C -\frac{1}{2}y^2\mathrm{d}x+0\mathrm{d}y
\\&=&\frac{1}{A}\oint_C -\frac{1}{2}y^2\mathrm{d}x
\\&=&-\frac{1}{2A}\oint_C y^2\mathrm{d}x
\end{eqnarray}
$$
X座標同様に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．
$$
\begin{eqnarray}
Y_G&=&-\frac{1}{2A}\oint_C y^2\mathrm{d}x
\\&=&-\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} y^2 \mathrm{d}x
\end{eqnarray}
$$
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．
$$
\begin{eqnarray}
y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)
\;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p)
\\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)
\end{eqnarray}
$$
$$
\begin{eqnarray}
\\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1}
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}{\color{red}+x_{k-1}y_{k-1}}+x_{k-1}y_k{\color{red}-x_{k-1}y_{k-1}}}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}}
\end{eqnarray}
$$
$$
\begin{eqnarray}
y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)
\\y&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)+y_{k-1}
\\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{y_k-y_{k-1}}{x_k-x_{k-1}}(-x_{k-1})+y_{k-1}
\\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{(-x_{k-1})(y_k-y_{k-1})+(x_k-x_{k-1})y_{k-1}}{x_k-x_{k-1}}
\\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}y_k{\color{red}+x_{k-1}y_{k-1}}+x_ky_{k-1}{\color{red}-x_{k-1}y_{k-1}}}{x_k-x_{k-1}}
\\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}y_k+x_ky_{k-1}}{x_k-x_{k-1}}
\\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}}
\end{eqnarray}
$$
区間毎の積分より重心を求める
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し積分を行う．まずX座標\(X_G\)について進める．
$$
\begin{eqnarray}
X_G&=&\frac{1}{2A}\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} x^2 \mathrm{d}y
\\&=&\frac{1}{2A}\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)^2 \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}}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\int^{y_k}_{y_{k-1}} \left\{
\left( \frac{x_k - x_{k-1}}{y_k - y_{k-1}}y \right)^2
+ 2 \left( \frac{x_k - x_{k-1}}{y_k - y_{k-1}}y \right) \left( \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right)
+ \left( \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right)^2
\right\} \mathrm{d}y
\;\cdots\;x=(A+B)^2=A^2+2AB+B^2
\\&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\int^{y_k}_{y_{k-1}} \left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}y\right)^2 \mathrm{d}y
+ \int^{y_k}_{y_{k-1}} 2\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}y\right)\left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right)\mathrm{d}y
+ \int^{y_k}_{y_{k-1}} \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \mathrm{d}y
\right\}
\;\cdots\;\int_X(A(x)+B(x)+C(x))\mathrm{d}x=\int_XA(x)\mathrm{d}x+\int_XB(x)\mathrm{d}x+\int_XC(x)\mathrm{d}x
\\&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)^2 \int^{y_k}_{y_{k-1}} y^2 \mathrm{d}y
+ 2\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)\left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right) \int^{y_k}_{y_{k-1}} y \mathrm{d}y
+ \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \int^{y_k}_{y_{k-1}} \mathrm{d}y
\right\}
\;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x
\\&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)^2 \left[\frac{1}{3}y^3\right]^{y_k}_{y_{k-1}}
+ 2\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)\left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right) \left[\frac{1}{2}y^2\right]^{y_k}_{y_{k-1}}
+ \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \left[y\right]^{y_k}_{y_{k-1}}
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\frac{1}{3}\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)^2 \left[y^3\right]^{y_k}_{y_{k-1}}
+ \frac{1}{2}2\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)\left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right) \left[y^2\right]^{y_k}_{y_{k-1}}
+ \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \left[y\right]^{y_k}_{y_{k-1}}
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\frac{1}{3}\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)^2\left(y_k^3-y_{k-1}^3\right)
+ \left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right) \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right)\left(y_k^2-y_{k-1}^2\right)
+ \left( \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \left(y_k-y_{k-1}\right)
\right\}
\end{eqnarray}
$$
\(\left\{\right\}\)の中をまとめる．
$$
\begin{eqnarray}
X_G&=&\frac{1}{2A}\sum_{k=1}^n\left\{
\frac{1}{3}\left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right)^2\left(y_k^3-y_{k-1}^3\right)
+ \left(\frac{x_k - x_{k-1}}{y_k - y_{k-1}}\right) \left(\frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}} \right)\left(y_k^2-y_{k-1}^2\right)
+ \left( \frac{x_{k-1}y_k - x_ky_{k-1}}{y_k - y_{k-1}}\right)^2 \left(y_k-y_{k-1}\right)
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{(y_k - y_{k-1})^2}
\left\{
\frac{1}{3}\left(x_k - x_{k-1}\right)^2\left(y_k^3-y_{k-1}^3\right)
+ \left(x_k - x_{k-1}\right) \left(x_{k-1}y_k - x_ky_{k-1} \right)\left(y_k^2-y_{k-1}^2\right)
+ \left(x_{k-1}y_k - x_ky_{k-1}\right)^2 \left(y_k-y_{k-1}\right)
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})^2}
\left\{
\left(x_k - x_{k-1}\right)^2\left(y_k-y_{k-1}\right)\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)
+ 3\left(x_k - x_{k-1}\right) \left(x_{k-1}y_k - x_ky_{k-1} \right)\left(y_k-y_{k-1}\right)\left(y_k+y_{k-1}\right)
+ 3\left(x_{k-1}y_k - x_ky_{k-1}\right)^2 \left(y_k-y_{k-1}\right)
\right\}
\\&&\;\cdots\;A^3-B^3=(A-B)(A^2+AB+B^2),\;A^2-B^2=(A-B)(A+B)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{\left(y_k-y_{k-1}\right)}{3(y_k - y_{k-1})^2}
\left\{
\left(x_k - x_{k-1}\right)^2\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)
+ 3\left(x_k - x_{k-1}\right) \left(x_{k-1}y_k - x_ky_{k-1} \right)\left(y_k+y_{k-1}\right)
+ 3\left(x_{k-1}y_k - x_ky_{k-1}\right)^2
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left\{
\left(x_k - x_{k-1}\right)^2\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)
+ 3\left(x_{k-1}y_k - x_ky_{k-1}\right)\left\{
\left(x_k - x_{k-1}\right)\left(y_k+y_{k-1}\right)
+ \left(x_{k-1}y_k - x_ky_{k-1}\right)
\right\}
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left\{
\left(x_k - x_{k-1}\right)^2\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)
+ 3\left(x_{k-1}y_k - x_ky_{k-1}\right)\left(x_k y_k-x_{k-1}y_{k-1}\right)
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left\{
\left(x_k^2 -2x_kx_{k-1}+ x_{k-1}^2\right)\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)
+ 3\left(x_{k-1}y_k - x_ky_{k-1}\right)\left(x_k y_k-x_{k-1}y_{k-1}\right)
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left(
x_k^2y_k^2+x_k^2y_ky_{k-1}+x_k^2y_{k-1}^2-2x_kx_{k-1}y_k^2-2x_kx_{k-1}y_ky_{k-1}-2x_kx_{k-1}y_{k-1}^2+x_{k-1}^2y_k^2+ x_{k-1}^2y_ky_{k-1}+ x_{k-1}^2y_{k-1}^2
+3x_kx_{k-1}y_k^2-3x_{k-1}^2y_ky_{k-1}-3x_k^2y_ky_{k-1}+3x_kx_{k-1}y_{k-1}^2
\right)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left(
x_k^2y_k^2
+x_k^2y_{k-1}^2
+x_{k-1}^2y_k^2
+x_{k-1}^2y_{k-1}^2
+ x_kx_{k-1}y_k^2
+ x_kx_{k-1}y_{k-1}^2
-2x_k^2y_ky_{k-1}
-2x_{k-1}^2y_ky_{k-1}
-2x_kx_{k-1}y_ky_{k-1}
\right)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left(
x_k^2(y_k^2+y_{k-1}^2)
+x_{k-1}^2(y_k^2+y_{k-1}^2)
+x_kx_{k-1}(y_k^2+y_{k-1}^2)
-2y_ky_{k-1}(x_k^2+x_kx_{k-1}+x_{k-1}^2)
\right)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left\{
\left(
x_k^2 +x_kx_{k-1} +x_{k-1}^2
\right)
\left(
y_k^2 +y_{k-1}^2
\right)
-2y_ky_{k-1}
\left(
x_k^2 +x_kx_{k-1} +x_{k-1}^2
\right)
\right\}
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left(
x_k^2 +x_kx_{k-1} +x_{k-1}^2
\right)
\left(
y_k^2 -2y_ky_{k-1}+y_{k-1}^2
\right)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3(y_k - y_{k-1})}
\left(
x_k^2 +x_kx_{k-1} +x_{k-1}^2
\right)
\left(
y_k -y_{k-1}
\right)^2
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{y_k -y_{k-1}}{3(y_k - y_{k-1})}
\left(
x_k^2 +x_kx_{k-1} +x_{k-1}^2
\right)
\left(
y_k -y_{k-1}
\right)
\\&=&\frac{1}{2A}\sum_{k=1}^n
\frac{1}{3}
\left(x_k^2 +x_kx_{k-1} +x_{k-1}^2\right)
\left(y_k -y_{k-1}\right)
\\&=&\frac{1}{2A}\frac{1}{3}\sum_{k=1}^n
\left(x_k^2 +x_kx_{k-1} +x_{k-1}^2\right)
\left(y_k -y_{k-1}\right)
\\&=&\frac{1}{6A}\sum_{k=1}^n
\left(x_k^2 +x_kx_{k-1} +x_{k-1}^2\right)
\left(y_k -y_{k-1}\right)
\end{eqnarray}
$$
多角形の頂点座標から，その重心のX座標\(X_G\)が求められた．
同様に重心のY座標\(Y_G\)は以下のようになる．
$$
\begin{eqnarray}
Y_G&=&-\frac{1}{2A}\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} y^2 \mathrm{d}y
\\&=&-\frac{1}{2A}\sum_{k=1}^n \int^{x_k}_{x_{k-1}} \left(
\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}}
\right)^2 \mathrm{d}y
\;\cdots\;y=\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}}
\\&=&-\frac{1}{6A}\sum_{k=1}^n
\left(y_k^2 +y_ky_{k-1} +y_{k-1}^2\right)
\left(x_k -x_{k-1}\right)
\end{eqnarray}
$$
例:正四角形
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4=x_0, y_4=y_0)&=&
(0, 0), (1, 0), (1, 1), (0, 1), (0, 0)
\\A&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post.html}{1}
\end{eqnarray}
$$
$$
\begin{eqnarray}
X_G&=&\frac{1}{6A}\sum_{k=1}^n
\left(x_k^2+x_kx_{k-1}+x_{k-1}^2\right)\left(y_k-y_{k-1}\right)
\\&=&\frac{1}{6\cdot1}\left\{
\left(1^2+1\cdot 0+0^2\right)\left(0-0\right)
+\left(1^2+1\cdot 1+1^2\right)\left(1-0\right)
+\left(0^2+0\cdot 1+1^2\right)\left(1-1\right)
+\left(0^2+0\cdot 0+0^2\right)\left(0-1\right)
\right\}
\\&=&\frac{1}{6}\left(
0+3+0+0
\right)
\\&=&\frac{1}{6}3
\\&=&\frac{1}{2}
\end{eqnarray}
$$
$$
\begin{eqnarray}
Y_G&=&-\frac{1}{6A}\sum_{k=1}^n
\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)\left(x_k-x_{k-1}\right)
\\&=&-\frac{1}{6\cdot1}\left\{
\left(0^2+0\cdot 0+0^2\right)\left(1-0\right)
+\left(1^2+1\cdot 0+0^2\right)\left(1-1\right)
+\left(1^2+1\cdot 1+1^2\right)\left(0-1\right)
+\left(0^2+0\cdot 1+1^2\right)\left(0-0\right)
\right\}
\\&=&-\frac{1}{6}\left(
0+0-3+0
\right)
\\&=&-\frac{1}{6}(-3)
\\&=&\frac{1}{2}
\end{eqnarray}
$$
よって重心は以下の点になる．
$$
\begin{eqnarray}
(X_G,Y_G)&=&(\frac{1}{2},\frac{1}{2})
\end{eqnarray}
$$
例:二等辺三角形
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1), (x_2, y_2), (x_3=x_0, y_3=y_0)&=&
(0, 0), (1, 0), (\frac{1}{2}, 1), (0, 0)
\\A&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post.html}{\frac{1}{2}}
\end{eqnarray}
$$
$$
\begin{eqnarray}
X_G&=&\frac{1}{6A}\sum_{k=1}^n
\left(x_k^2+x_kx_{k-1}+x_{k-1}^2\right)\left(y_k-y_{k-1}\right)
\\&=&\frac{1}{6\cdot\frac{1}{2}}\left[
\left(1^2+1\cdot 0+0^2\right)\left(0-0\right)
+\left\{\left(\frac{1}{2}\right)^2+\frac{1}{2}\cdot 1+1^2\right\}\left(1-0\right)
+\left\{0^2+0\cdot \frac{1}{2}+\left(\frac{1}{2}\right)^2\right\}\left(0-1\right)
\right]
\\&=&\frac{1}{3}\left(
0+\frac{7}{4}-\frac{1}{4}
\right)
\\&=&\frac{1}{3}\frac{6}{4}
\\&=&\frac{1}{2}
\end{eqnarray}
$$
$$
\begin{eqnarray}
Y_G&=&-\frac{1}{6A}\sum_{k=1}^n
\left(y_k^2+y_ky_{k-1}+y_{k-1}^2\right)\left(x_k-x_{k-1}\right)
\\&=&-\frac{1}{6\cdot\frac{1}{2}}\left\{
\left(0^2+0\cdot 0+0^2\right)\left(1-0\right)
+\left(1^2+1\cdot 0+0^2\right)\left(\frac{1}{2}-1\right)
+\left(0^2+0\cdot 1+1^2\right)\left(0-\frac{1}{2}\right)
\right\}
\\&=&-\frac{1}{3}\left(
0-\frac{1}{2}-\frac{1}{2}
\right)
\\&=&-\frac{1}{3}(-1)
\\&=&\frac{1}{3}
\end{eqnarray}
$$
よって重心は以下の点になる．
$$
\begin{eqnarray}
(X_G,Y_G)&=&(\frac{1}{2},\frac{1}{3})
\end{eqnarray}
$$
