多角形の頂点座標から面積を求める (グリーンの定理を用いて面積分を周回積分にして求める)

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。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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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 $$\begin{array}{rcl}A& =& {\iint }_{D}1\mathrm{d}x\mathrm{d}y\end{array}$$ 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 $$\begin{array}{rcl}Q(x,y)& =& x\\ P(x,y)& =& 0\end{array}$$ とおく(それぞれ一階微分可能で連続($C^1$級凾数))．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& 1-0\\ & =& 1\end{array}$$ 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． $$\begin{array}{rcl}A& =& {\iint }_{D}1\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& {\oint }_{C}0\mathrm{d}x+x\mathrm{d}y\\ & =& {\oint }_{C}x\mathrm{d}y\end{array}$$ 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． $$\begin{array}{r}(x_0,y_0),(x_1,y_1),(x_2,y_2),\cdots ,(x_n,y_n)=(x_0,y_0)\end{array}$$ 

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． $$\begin{array}{rcl}A& =& {\oint }_{C}x\mathrm{d}y\\ & =& \sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}x\mathrm{d}y\end{array}$$

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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． $$\begin{array}{rcl}{y}_{-}{y}_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\cdots \mathrm{傾}\mathrm{き}\theta \mathrm{の}\mathrm{直}\mathrm{線}\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{原}\mathrm{点}\mathrm{を}，\mathrm{点}(p,q)\mathrm{へ}\mathrm{移}\mathrm{動}\mathrm{し}\mathrm{た}\mathrm{式}\mathrm{と}\mathrm{同}\mathrm{じ}．(y-q)=\theta (x-p)\\ \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})& =& x-x_{k-1}\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})+x_{k-1}\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{(x_k-x_{k-1})(-y_{k-1})+x_{k-1}(y_k-y_{k-1})}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_k}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\end{array}$$

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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{array}{rcl}A& =& \sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}x\mathrm{d}y\\ & =& \sum _{k=1}^n{\int }_{y_{k-1}}^{y_k}(\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}})\mathrm{d}y\cdots x=\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \sum _{k=1}^n\{{\int }_{y_{k-1}}^{y_k}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y\mathrm{d}y+{\int }_{y_{k-1}}^{y_k}\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\mathrm{d}y\}\cdots {\int }_X(A(x)+B(x))\mathrm{d}x={\int }_{X}A(x)\mathrm{d}x+{\int }_{X}B(x)\mathrm{d}x\\ & =& \sum _{k=1}^n\{\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\int }_{y_{k-1}}^{y_k}y\mathrm{d}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\int }_{y_{k-1}}^{y_k}\mathrm{d}y\}\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\\ & =& \sum _{k=1}^n\{\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\left[\frac{1}{2}{y}^2\right]}_{y_{k-1}}^{y_k}+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}{\left[y^2\right]}_{y_{k-1}}^{y_k}+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}{\left[y\right]}_{y_{k-1}}^{y_k}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y_k^2-y_{k-1}^2)+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}(y_k-y_{k-1})\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y_k+y_{k-1})(y_k-y_{k-1})+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}(y_k-y_{k-1})\}\cdots A^2-B^2=(A+B)(A-B)\\ & =& \sum _{k=1}^n\{\frac{1}{2}(x_k-x_{k-1})(y_k+y_{k-1})+x_{k-1}{y}_k-x_k{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}(x_k{y}_k+x_k{y}_{k-1}-x_{k-1}{y}_k-x_{k-1}{y}_{k-1})+x_{k-1}{y}_k-x_k{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}{x}_k{y}_k+(\frac{1}{2}-1)x_k{y}_{k-1}+(-\frac{1}{2}+1)x_{k-1}{y}_k-\frac{1}{2}{x}_{k-1}{y}_{k-1}\}\\ & =& \sum _{k=1}^n\{\frac{1}{2}{x}_k{y}_k-\frac{1}{2}{x}_k{y}_{k-1}+\frac{1}{2}{x}_{k-1}{y}_k-\frac{1}{2}{x}_{k-1}{y}_{k-1}\}\\ & =& \sum _{k=1}^n\left\{\frac{1}{2}(x_k{y}_k-x_k{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1})\right\}\\ & =& \sum _{k=1}^n\left[\frac{1}{2}\{x_k(y_k-y_{k-1})+x_{k-1}(y_k-y_{k-1})\}\right]\\ & =& \sum _{k=1}^n\frac{1}{2}(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\end{array}$$ 多角形の頂点座標から，その面積が求められた．

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例:正四角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4=x_0,y_4=y_0)& =& (0,0),(1,0),(1,1),(0,1),(0,0)\end{array}$$ $$\begin{array}{rcl}A& =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\{(x_1+x_0)(y_1-y_0)+(x_2+x_1)(y_2-y_1)+(x_3+x_2)(y_3-y_2)+(x_4+x_3)(y_4-y_3)\}\\ & =& \frac{1}{2}\{(1+0)(0-0)+(1+1)(1-0)+(0+1)(1-1)+(0+0)(0-1)\}\\ & =& \frac{1}{2}(0+2+0+0)\\ & =& \frac{1}{2}2\\ & =& 1\end{array}$$

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例:二等辺三角形

$$\begin{array}{rcl}(x_0,y_0),(x_1,y_1),(x_2,y_2),(x_3=x_0,y_3=y_0)& =& (0,0),(1,0),(\frac{1}{2},1),(0,0)\end{array}$$ $$\begin{array}{rcl}A& =& \frac{1}{2}\sum _{k=1}^n(x_k+x_{k-1})(y_k-y_{k-1})\\ & =& \frac{1}{2}\{(x_1+x_0)(y_1-y_0)+(x_2+x_1)(y_2-y_1)+(x_3+x_2)(y_3-y_2)\}\\ & =& \frac{1}{2}\{(1+0)(0-0)+(\frac{1}{2}+1)(1-0)+(0+\frac{1}{2})(0-1)\}\\ & =& \frac{1}{2}(0+\frac{3}{2}-\frac{1}{2})\\ & =& \frac{1}{2}1\\ & =& \frac{1}{2}\end{array}$$