多角形D上の点から任意の点への距離の二乗の平均値(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。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

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面積分を周回積分へ変形し，区間毎の線積分に展開する

点\(x,y\)から点\((p,q)\)までの距離の二乗は $$ \begin{eqnarray} \langle r^2 \rangle&=&\left\{\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right\}^2 \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ であり，多角形D上の点(\(x,y\))から任意の点(\(p,q\))への距離の二乗の平均値\(\langle r^2 \rangle\)は， $$ \begin{eqnarray} \langle r^2 \rangle&=&\iint_D \rho(x,y)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\};\mathrm{d}x\mathrm{d}y \\&=&\iint_D \frac{1}{A}\left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積) \\&=&\frac{1}{A}\iint_D \left(x-p\right)^2+\left(y-q\right)^2\;\mathrm{d}x\mathrm{d}y \;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x \end{eqnarray} $$ である．

これをグリーンの定理の式で満たすために例えば $$ \begin{eqnarray} Q(x, y)&=&\int \left(x-p\right)^2\;\mathrm{d}x \\P(x, y)&=&-\int \left(y-q\right)^2\;\mathrm{d}y \end{eqnarray} $$ とおく．これは $$ \begin{eqnarray} \frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y} &=& \int \left(x-p\right)^2\;\mathrm{d}x - \left\{-\int \left(y-q\right)^2\;\mathrm{d}y\right\} \\&=&\left(x-p\right)^2+\left(y-q\right)^2 \end{eqnarray} $$ 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\(P,Q\)を用いて線積分側の 被積分凾数を求めると以下のようになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\int_D \rho\left(x, y\right)r\left(x,y;p, q\right)^2 \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \rho\left(x, y\right)\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\int_D \frac{1}{A}\left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\int_D \left\{\left(x-p\right)^2+\left(y-q\right)^2\right\} \;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\left[ \int_D \left[ \frac{\partial }{\partial x} \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} - \frac{\partial }{\partial y} \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \right] \mathrm{d}x\mathrm{d}y \right] \\&=&\frac{1}{A}\left[ \oint_C \left\{ \int \left(x-p\right)^2\;\mathrm{d}x \right\} \mathrm{d}y + \left\{ -\int \left(y-q\right)^2\;\mathrm{d}y \right\} \mathrm{d}x \right] \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y - \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \oint_C \frac{1}{3}\left(x-p\right)^3\;\mathrm{d}y -\oint_C \frac{1}{3}\left(y-q\right)^3\;\mathrm{d}x \right\} \\&=&\frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \end{eqnarray} $$

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\(\left\{\right\}\)内の第一項 区間毎の線積分

$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=& \frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{\left(\epsilon_k y+\zeta_k\right)-p\right\}^3\;\mathrm{d}y \;\cdots\;x=\epsilon_k y+\zeta_k 多角形の区間(頂点k-1からk)毎の直線の式 \\&=&\frac{1}{3}\sum_{k=1}^n \int_{y_{k-1}}^{y_k} \left\{ \left(\epsilon_k y+\zeta_k\right)^3 -3\left(\epsilon_k y+\zeta_k\right)^2 p +3\left(\epsilon_k y+\zeta_k\right) p^2 -p^3 \right\}\;\mathrm{d}y \;\cdots\;(A-B)^3=A^3-3A^2B+3AB^2-B^3 \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^4 -3\frac{1}{3\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^3 p +3\frac{1}{2\epsilon_k}\left(\epsilon_k y+\zeta_k\right)^2 p^2 -y p^3 \right]_{y_{k-1}}^{y_k} \;\cdots\;\int cf(x) \mathrm{d}x= c\int f(x) \mathrm{d}x ,\;\int_a^b x^a \mathrm{d}x= \left[\frac{1}{a+1}x^{a+1}\right]_a^b \\&=&\frac{1}{3}\sum_{k=1} \left[ \frac{1}{4\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^4 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^4 \right\} -\frac{1}{\epsilon_k}\left\{ \left(\epsilon_k y_k+\zeta_k\right)^3 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^3 \right\} p +\frac{3}{2\epsilon_k}\left\{ \left(\epsilon_k y_{k}+\zeta_k\right)^2 -\left(\epsilon_k y_{k-1}+\zeta_k\right)^2 \right\} p^2 -\left( y_k - y_{k-1} \right) p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}^4-y_{k-1}^4) + 4 \epsilon_k^3 \zeta_k (y_{k}^3-y_{k-1}^3) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}^2-y_{k-1}^2) + 4 \epsilon_k \zeta_k^3 (y_{k} -y_{k-1} ) \right\} -\frac{1}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^3-y_{k-1}^3) + 3 \epsilon_k^2 \zeta_k (y_{k}^2-y_{k-1}^2) + 3 \epsilon_k \zeta_k^2 (y_{k} -y_{k-1} ) \right\} p +\frac{3}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}^2-y_{k-1}^2) + 2 \epsilon_k \zeta_k (y_{k} -y_{k-1} ) \right\} p^2 -\left( y_k -y_{k-1} \right) p^3 \right] \\&&\;\cdots\;(a x + b)^4-(a y + b)^4 =a^4 x^4 - a^4 y^4 + 4 a^3 b x^3 - 4 a^3 b y^3 + 6 a^2 b^2 x^2 - 6 a^2 b^2 y^2 + 4 a b^3 x - 4 a b^3 y =a^4 (x^4 - y^4) + 4 a^3 b (x^3 - y^3) + 6 a^2 b^2 (x^2 - y^2) + 4 a b^3 (x - y) \\&&\;\cdots\;(a x + b)^3-(a y + b)^3 =a^3 x^3 - a^3 y^3 + 3 a^2 b x^2 - 3 a^2 b y^2 + 3 a b^2 x - 3 a b^2 y =a^3 (x^3 - y^3) + 3 a^2 b (x^2 - y^2) + 3 a b^2 (x - y) \\&&\;\cdots\;(a x + b)^2-(a y + b)^2 =a^2 x^2 - a^2 y^2 + 2 a b x - 2 a b y =a^2 (x^2 - y^2) + 2 a b (x - y) \\&=&\frac{1}{3}\sum_{k=1}^n \left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k}-y_{k-1})(y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 (y_{k}-y_{k-1}) \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}-y_{k-1})(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 (y_{k}-y_{k-1}) \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k}-y_{k-1})(y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k (y_{k}-y_{k-1}) \right\} -p^3\left( y_k -y_{k-1} \right) \right] \\&&\;\cdots\;(a^4-b^4)=(a-b)(a+b)(a^2+b^2) \\&&\;\cdots\;(a^3-b^3)=(a-b)(a^2+ab+b^2) \\&&\;\cdots\;(a^2-b^2)=(a-b)(a+b) \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4\epsilon_k}\left\{ \epsilon_k^4 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^3 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k^2 \zeta_k^2 (y_{k} +y_{k-1}) + 4 \epsilon_k \zeta_k^3 \right\} -\frac{p}{\epsilon_k}\left\{ \epsilon_k^3 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k^2 \zeta_k (y_{k} +y_{k-1}) + 3 \epsilon_k \zeta_k^2 \right\} +\frac{3p^2}{2\epsilon_k}\left\{ \epsilon_k^2 (y_{k} +y_{k-1}) + 2 \epsilon_k \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \end{eqnarray} $$

\(\epsilon_k,\zeta_k\)の代入と整理

$$ \begin{eqnarray} \epsilon_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{x_{k}-x_{k-1}}{y_{k}-y_{k-1}}} \\ \zeta_k&=&\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{\frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1})}} \\ \\&& \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \\&=& \frac{(x_{k}-x_{k-1})^3}{(y_{k}-y_{k-1})^3} (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \frac{(x_{k}-x_{k-1})}{(y_{k}-y_{k-1})} \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k} +y_{k-1}) + 4 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^3}{(y_{k}-y_{k-1})^3} \\&=&\frac{1}{(y_{k}-y_{k-1})^3} \left\{ (x_{k}-x_{k-1})^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 (x_{k}-x_{k-1})^2 (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1})^2 (y_{k} +y_{k-1}) + 4 (x_{k-1}y_{k}-x_{k}y_{k-1})^3 \right\} \\&=& (x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2) \\ \\&& \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \\&=& \frac{(x_{k}-x_{k-1})^2}{(y_{k}-y_{k-1})^2} (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \frac{(x_{k}-x_{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}) + 3 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1})^2}{(y_{k}-y_{k-1})^2} \\&=&\frac{1}{(y_{k}-y_{k-1})^2} \left\{ (x_{k}-x_{k-1})^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 (x_{k}-x_{k-1}) (x_{k-1}y_{k}-x_{k}y_{k-1}) (y_{k} +y_{k-1}) + 3 (x_{k-1}y_{k}-x_{k}y_{k-1})^2 \right\} \\&=&x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2 \\ \\&& \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \\&=& \frac{(x_{k}-x_{k-1}) }{(y_{k}-y_{k-1}) } (y_{k} +y_{k-1}) + 2 \frac{(x_{k-1}y_{k}-x_{k}y_{k-1}) }{(y_{k}-y_{k-1}) } \\&=&\frac{1}{(y_{k}-y_{k-1})} \left\{ (x_{k}-x_{k-1}) (y_{k} +y_{k-1}) + 2 (x_{k-1}y_{k}-x_{k}y_{k-1}) \right\} \\&=&x_{k}+x_{k-1} \end{eqnarray} $$

\(\left\{\right\}\)内の第一項　整理後

$$ \begin{eqnarray} \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y &=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{ \epsilon_k^3 (y_{k} +y_{k-1})(y_{k}^2+y_{k-1}^2) + 4 \epsilon_k^2 \zeta_k (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 6 \epsilon_k \zeta_k^2 (y_{k} +y_{k-1}) + 4 \zeta_k^3 \right\} -p\left\{ \epsilon_k^2 (y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2) + 3 \epsilon_k \zeta_k (y_{k} +y_{k-1}) + 3 \zeta_k^2 \right\} +\frac{3p^2}{2}\left\{ \epsilon_k (y_{k} +y_{k-1}) + 2 \zeta_k \right\} -p^3 \right] \\&=&\frac{1}{3}\sum_{k=1}^n (y_{k}-y_{k-1})\left[ \frac{1}{4}\left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -p\left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{3p^2}{2}\left(x_{k}+x_{k-1}\right) -p^3 \right] \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) -\frac{p^3}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \\&=& \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \;\cdots\;\sum_{k=1}^n (y_{k}-y_{k-1}) = 0\;(周回積分なのでy軸の移動量の和は一周したら元に戻る) \end{eqnarray} $$

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\(\left\{\right\}\)内の第二項　整理後

$$ \begin{eqnarray} -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x &=& -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \end{eqnarray} $$

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項毎に計算したものをまとめる

$$ \begin{eqnarray} \langle r^2 \rangle &=& \frac{1}{A}\left\{ \frac{1}{3}\oint_C \left(x-p\right)^3\;\mathrm{d}y -\frac{1}{3}\oint_C \left(y-q\right)^3\;\mathrm{d}x \right\} \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{p}{3} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) +\frac{p^2}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) \right] -\left[ \frac{1}{12} \sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -\frac{q}{3} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) +\frac{q^2}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right) \right] \right] \\&=& \frac{1}{A}\left[ \left[ \frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} -\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} -2p \frac{1}{6} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right) -2q \left\{-\frac{1}{6} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right) \right\} +p^2\frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right) +q^2\left\{-\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\right\} \right] \right] \\&=& \frac{1}{A}\left[ \lambda+\mu -2p A X_G-2q A Y_G +p^2 A+q^2 A \right] \\&&\;\cdots\;\lambda=\frac{1}{12} \sum_{k=1}^n (y_{k}-y_{k-1}) \left\{(x_{k}+x_{k-1})(x_{k}^2+x_{k-1}^2)\right\} \\&&\;\cdots\;\mu=-\frac{1}{12}\sum_{k=1}^n (x_{k}-x_{k-1}) \left\{(y_{k}+y_{k-1})(y_{k}^2+y_{k-1}^2)\right\} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{X_G=\frac{1}{6A} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}^2+x_{k}x_{k-1}+x_{k-1}^2\right)\;(重心X軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post_24.html}{Y_G=-\frac{1}{6A} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}^2+y_{k}y_{k-1}+y_{k-1}^2\right)\;(重心Y軸位置)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/blog-post.html}{A= \frac{1}{2} \sum_{k=1}^n (y_{k}-y_{k-1}) \left(x_{k}+x_{k-1}\right)\;(面積)} \\&&\;\cdots\;A= -\frac{1}{2} \sum_{k=1}^n (x_{k}-x_{k-1}) \left(y_{k}+y_{k-1}\right)\;(面積) \\&=& \frac{\lambda+\mu}{A} -2p X_G -2q Y_G +p^2 +q^2 \\&=& \frac{\lambda+\mu}{A} {\color{red}+X_G^2} -2p X_G +p^2 {\color{red}+Y_G^2} -2q Y_G +q^2 {\color{red}-X_G^2} {\color{red}-Y_G^2} \\&=& \frac{\lambda+\mu}{A} +\left(p-X_G\right)^2 +\left(q-Y_G\right)^2 -X_G^2 -Y_G^2 \end{eqnarray} $$ よって\(\langle r^2 \rangle\)は多角形によって事前に決定される\(A,X_G,Y_G,\lambda,\mu\)と，点\((p,q)\)が与えられることで決まる．

\(\langle r^2 \rangle\)の最小値

\(\langle r^2 \rangle\)の最小値となる点\((u, v)\)は，\(\left(p-X_G\right),\left(p-X_G\right)\)が0となる点\((X_G, Y_G)\)である．

また，点\((X_G, Y_G)\)から\(\langle r^2 \rangle\)の等しい点\((p, q)\)は，以下の式を満たすことになる． $$ \begin{eqnarray} \langle r^2 \rangle&=&\frac{\lambda+\mu}{A}+\left(p-X_G\right)^2+\left(q-Y_G\right)^2-X_G^2-Y_G^2 \\\left(p-X_G\right)^2+\left(q-Y_G\right)^2&=&\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A} \\\sqrt{\left(p-X_G\right)^2+\left(q-Y_G\right)^2}&=&\sqrt{\langle r^2 \rangle+X_G^2+Y_G^2-\frac{\lambda+\mu}{A}} \end{eqnarray} $$

多角形D上の点から任意の点への距離の平均値 (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。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

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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

点\((x,y)\)から点\((p,q)\)までの距離は $$ \begin{eqnarray} r(x, y; p, q)&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \end{eqnarray} $$ であり，多角形D上の点\(x,y\)から任意の点\(p,q\)への距離の平均値\(\bar{r}\)は， $$ \begin{eqnarray} \bar{r}&=&\iint_D \rho(x,y)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \\&=&\iint_D \frac{1}{A}\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積) \\&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x \end{eqnarray} $$ である．

これをグリーンの定理の式で満たすために例えば $$ \begin{eqnarray} Q(x, y)&=&\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x \\P(x, y)&=&0 \end{eqnarray} $$ とおく．これは $$ \begin{eqnarray} \frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}-0 \\&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \end{eqnarray} $$ 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． $$ \begin{eqnarray} \bar{r}&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y \\&=&\frac{1}{A}\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \\&=&\frac{1}{A}\oint_C 0\mathrm{d}x+\left(\int_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y \\&=&\frac{1}{A}\oint_C \left(\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y \\&=&\frac{1}{A}\oint_C \left(\int \sqrt{u^2+v^2}\;\mathrm{d}u\right)\mathrm{d}y \;\cdots\;u=x-p, v=y-q \\&=&\frac{1}{A}\oint_C \frac{1}{2}\left\{u\sqrt{u^2+v^2}+v^2\ln{\left|u+\sqrt{u^2+v^2}\right|}\right\}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2}\;\mathrm{d}x=\frac{1}{2}\left\{x\sqrt{x^2+a^2}+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}\right\} + 積分定数} \\&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \end{eqnarray} $$ 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． $$ \begin{eqnarray} \bar{r}&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \\&=&\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} +\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y \\&=&\frac{1}{2A} \sum_{k=1}^n\left\{ \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y +\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}+ I_{2k}\right\} \\&&\;\cdots\;I_{1k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \\&&\;\cdots\;I_{2k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \end{eqnarray} $$

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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． $$ \begin{eqnarray} y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right) \;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p) \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \end{eqnarray} $$ $$ \begin{eqnarray} \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1} \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y +\frac{ -x_ky_{k-1} +x_{k-1}y_{k-1} +x_{k-1}y_k -x_{k-1}y_{k-1} }{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}} \\&=&\epsilon_k y+\zeta_k \end{eqnarray} $$

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区間毎の積分より距離の平均値\(\bar{r}\)を求める　第一項\(I_{1k}\)

$$ \begin{eqnarray} I_{1k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \\&=&I_{1k}(y_k)-I_{1k}(y_{k-1}) \\I_{1k}(y)&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right) \sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y \;\cdots\;x=\epsilon_k y+\zeta_k \\&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right) \sqrt{\left(\left(\epsilon_k y+\zeta_k\right)^2-2p\left(\epsilon_k y+\zeta_k\right)+p^2\right)+\left(y^2-2qy+q^2\right)} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\left(\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2\right)-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right) \sqrt{\left(\epsilon_k^2 +1 \right)y^2+2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\}y +\left(\zeta_k-p\right)^2+q^2} \;\mathrm{d}y \\&=&\int \left(\epsilon_k y+\zeta_k-p\right)\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \;\cdots\;\alpha_k=\epsilon_k^2 +1,\;\beta_k=2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\},\;\gamma_k=\left(\zeta_k-p\right)^2+q^2 \\&=& \int \epsilon_k y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \int \left(\zeta_k-p\right) \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y + \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } +C_0 \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } +C_1 \right\} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/xa-x2b-x-c.html}{\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x= \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C_0\;(C_0:積分定数)} \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/a-x2b-x-c.html}{\int \sqrt{a x^2+b x +c} \;\mathrm{d}x= \frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C_1\;(C_1:積分定数)} \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} +\epsilon_k C_0 +\left(\zeta_k-p\right) C_1 \\&=& \epsilon_k \left\{ \frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}} -\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} \\&&+ \left(\zeta_k-p\right) \left\{ \frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} +\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| } \right\} + C \;\cdots\;C=\epsilon_k C_0+\left(\zeta_k-p\right) C_1\;(C:積分定数) \end{eqnarray} $$

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区間毎の積分より距離の平均値\(\bar{r}\)を求める　第二項\(I_{2k}\)

$$ \begin{eqnarray} I_{2k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2\ln{\left|\left(\left(\epsilon_k y+\zeta_k\right)-p\right)+\sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left( \left( \epsilon_k y+\zeta_k \right)^2 -2p\left( \epsilon_k y+\zeta_k \right)+p^2 \right) +\left( y^2-2qy-q^2 \right) } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left( \epsilon_k^2 y^2 +2\epsilon_k\zeta_k y +\zeta_k^2 \right) -2p\epsilon_k y -2p\zeta_k +p^2 +y^2-2qy-q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \epsilon_k^2 y^2 +y^2 +2\epsilon_k\zeta_k y -2p\epsilon_k y -2qy +\zeta_k^2 -2p\zeta_k +p^2 -q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{ \left(\epsilon_k^2+1\right) y^2 +2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\} y +\left(\zeta_k-p\right)^2 -q^2 } \right|} \;\mathrm{d}y \\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2 \ln{\left| \epsilon_k y+\zeta_k-p +\sqrt{\alpha_k y^2+\beta_k y+\gamma_k} \right|}\;\mathrm{d}y \;\cdots\;\alpha_k=\epsilon_k^2+1 ,\;\beta_k=2\left\{\epsilon_k \left(\zeta_k-p\right) -q\right\} ,\;\gamma_k=\left(\zeta_k-p\right)^2-q^2 \end{eqnarray} $$ →要:数値計算

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区間毎の積分より距離の平均値\(\bar{r}\)を求める　第一項\(I_{1k}\)+第二項\(I_{2k}\)

$$ \begin{eqnarray} \bar{r}&=&\frac{1}{2A} \sum_{k=1}^n\left\{ \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y +\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y \right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k} + I_{2k}\right\} \\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}(y_{k}) - I_{1k}(y_{k-1}) + I_{2k}\right\} \end{eqnarray} $$

x√(a x^2+b x +c)の積分

$$ \begin{eqnarray} \int x\sqrt{a x^2+b x +c} \;\mathrm{d}x &=&\int x\sqrt{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}+c} \;\mathrm{d}x \;\cdots\;平方完成 \\&&\;\cdots\;ax^2+bx+c=(\sqrt{a}x+\alpha)^2-\alpha^2+c \\&&\;\cdots\;(\sqrt{a}x+\alpha)^2=ax^2+2\sqrt{a}\alpha x+\alpha^2,\;2\sqrt{a}\alpha=bより\alpha=\frac{b}{2\sqrt{a}} \\&=&\int \frac{2\sqrt{a}u-b}{2a}\; \sqrt{u^2-\frac{b^2}{4a}+c} \;\frac{1}{\sqrt{a}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{a} x+\frac{b}{2\sqrt{a}} \\&&\;\cdots\;x=\frac{u-\frac{b}{2\sqrt{a}}}{\sqrt{a}} =\frac{\frac{2\sqrt{a}u-b}{2\sqrt{a}}}{\sqrt{a}} =\frac{2\sqrt{a}u-b}{2a} \\&&\;\cdots\;\frac{\mathrm{d}x}{\mathrm{d}u}=\frac{1}{\sqrt{a}},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u \\&=&\frac{1}{2a^{\frac{3}{2}}} \int \left(2\sqrt{a}u-b\right)\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{2a^{\frac{3}{2}}}\int 2\sqrt{a}u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{1}{2a^{\frac{3}{2}}} \int b\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{2\sqrt{a}}{2a^{\frac{3}{2}}}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \end{eqnarray} $$

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$$ \begin{eqnarray} \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=& \frac{1}{a}\int u\;\sqrt{s} \;\frac{1}{2u}\;\mathrm{d}s \;\cdots\;s=u^2-\frac{b^2}{4a}+c,\frac{\mathrm{d}s}{\mathrm{d}u}=2u,\mathrm{d}u=\frac{1}{2u}\;\mathrm{d}s \\&=& \frac{1}{2a}\int\sqrt{s}\;\mathrm{d}s \\&=& \frac{1}{2a}\left[\frac{2}{3}s^{\frac{3}{2}}+C_0\right]\;\cdots\;\int x^{\frac{1}{2}}\;\mathrm{d}x=\frac{2}{3}x^{\frac{3}{2}}+C_0\;(C_0:積分定数) \\&=& \frac{1}{3a}s^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(u^2-\frac{b^2}{4a}+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left\{\left(ax^2+bx+\frac{b^2}{4a}\right)-\frac{b^2}{4a}+c\right\}^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \end{eqnarray} $$

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$$ \begin{eqnarray} -\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u &=&-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}} ,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}} ,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t \\&&\;\cdots\;u^2+c-\frac{b^2}{4a} =\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right) =\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\} =\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right) \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right)\int \sqrt{t^2+1} \;\mathrm{d}t \\&=& - \frac{b}{2a^{\frac{3}{2}}} \left(c-\frac{b^2}{4a}\right) \left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right)\right\} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_1\right)\;(C_1:積分定数)} \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{bc}{4a^{\frac{3}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \left(\frac{b^3}{16a^{\frac{5}{2}}}-\frac{4abc}{16a^{\frac{5}{2}}}\right) \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_1\right) \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}} +\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_1 \right\} \\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| } +C_1 \right\} \\&=& \frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)} +\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\ln{\sqrt{4ac-b^2}} +C_1 \right\} \\&=&\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{\frac{4}{2}}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=&-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \end{eqnarray} $$

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$$ \begin{eqnarray} \int x\sqrt{a x^2+b x +c} \;\mathrm{d}x &=& \frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u - \frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&&\;\cdots\;\frac{1}{a}\int u\;\sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=\frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}+\frac{1}{2a}C_0 \\&&\;\cdots\;-\frac{b}{2a^{\frac{3}{2}}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u=-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1 \\&=& \frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}} -\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C \;\cdots\;C=\frac{1}{2a}C_0 -\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}C_1\;(C:積分定数) \end{eqnarray} $$

√(a x^2+b x +c)の積分

$$ \begin{eqnarray} \int \sqrt{a x^2+b x +c} \;\mathrm{d}x &=&\int \sqrt{\left(\sqrt{a} x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}+c} \;\mathrm{d}x \;\cdots\;平方完成 \\&&\;\cdots\;ax^2+bx+c=(\sqrt{a}x+\alpha)^2-\alpha^2+c \\&&\;\cdots\;(\sqrt{a}x+\alpha)^2=ax^2+2\sqrt{a}\alpha x+\alpha^2,\;2\sqrt{a}\alpha=bより\alpha=\frac{b}{2\sqrt{a}} \\&=&\int \sqrt{u^2-\frac{b^2}{4a}+c} \;\frac{1}{\sqrt{a}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{a} x+\frac{b}{2\sqrt{a}} \\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}x}=\sqrt{a},\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}u \\&=& \frac{1}{\sqrt{a}} \int \sqrt{u^2-\frac{b^2}{4a}+c} \;\mathrm{d}u \\&=& \frac{1}{\sqrt{a}} \int \sqrt{\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right)} \;\sqrt{c-\frac{b^2}{4a}}\;\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{\sqrt{c-\frac{b^2}{4a}}} ,\frac{\mathrm{d}t}{\mathrm{d}u}=\frac{1}{\sqrt{c-\frac{b^2}{4a}}} ,\mathrm{d}u=\sqrt{c-\frac{b^2}{4a}}\mathrm{d}t \\&&\;\cdots\;u^2+c-\frac{b^2}{4a} =\left(u^2+c-\frac{b^2}{4a}\right)\frac{c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\frac{u^2+c-\frac{b^2}{4a}}{c-\frac{b^2}{4a}} =\left(c-\frac{b^2}{4a}\right)\left(\frac{u^2}{c-\frac{b^2}{4a}}+1\right) =\left(c-\frac{b^2}{4a}\right)\left\{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1\right\} =\left(c-\frac{b^2}{4a}\right)\left(t^2+1\right) \\&=& \frac{4ac-b^2}{4a^{\frac{3}{2}}} \int \sqrt{t^2+1} \;\mathrm{d}t \\&=& \frac{4ac-b^2}{4a^{\frac{3}{2}}} \left\{\frac{1}{2}\left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right)\right\} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2} \;\mathrm{d}x = \frac{1}{2}\left(x\sqrt{x^2+a^2}+ a^2 \ln{ \left|x+\sqrt{x^2+a^2}\right| }+C_0\right)\;(C_0:積分定数)} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left(t\sqrt{t^2+1}+\ln{ \left|t+\sqrt{t^2+1}\right| }+C_0\right) \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left\{ \left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{u}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left\{ \left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1} +\ln{ \left|\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)+\sqrt{\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1}\right| }+C_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left\{ \frac{2ax+b}{\sqrt{4ac-b^2}}\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}} +\ln{ \left|\frac{2ax+b}{\sqrt{4ac-b^2}}+\sqrt{\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2}}\right| }+C_0 \right\} \\&&\;\cdots\;\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}=\frac{2ax+b}{\sqrt{4ac-b^2}} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2=\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)^2=\frac{4a^2x^2+4abx+b^2}{4ac-b^2} \\&&\;\cdots\;\left(\frac{\sqrt{a} x+\frac{b}{2\sqrt{a}}}{\sqrt{c-\frac{b^2}{4a}}}\right)^2+1=\frac{4a^2x^2+4abx+b^2}{4ac-b^2}+1=\frac{4a^2x^2+4abx+b^2+4ac-b^2}{4ac-b^2}=\frac{4a^2x^2+4abx+4ac}{4ac-b^2}=\frac{4a\left(ax^2+bx+c\right)}{4ac-b^2} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\ln{ \left|\frac{1}{\sqrt{4ac-b^2}}\left\{\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right\}\right| } +C_0 \right\} \\&=& \frac{4ac-b^2}{8a^{\frac{3}{2}}} \left\{ \frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{a\left(ax^2+bx+c\right)} +\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\ln{\sqrt{4ac-b^2}} +C_0 \right\} \\&=&\frac{4ac-b^2}{8a^{\frac{3}{2}}}\frac{2ax+b}{4ac-b^2}2\sqrt{a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } -\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \\&=&\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)} +\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| } +C \;\cdots\;C=-\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{\sqrt{4ac-b^2}}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}C_0 \;(C:積分定数) \end{eqnarray} $$

√(x^2+a^2)の積分

$$ \begin{eqnarray} \int \sqrt{x^2+a^2} \mathrm{d}x &=&\int \sqrt{a^2\tan^2{\left(\theta\right)}+a^2}\;a\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \;\cdots\; x=a\tan{\left(\theta\right)}, \frac{\mathrm{d}x}{\mathrm{d}\theta} =a\frac{\mathrm{d}}{\mathrm{d}\theta}\tan{\left(\theta\right)} =a\frac{1}{\cos^2{\left(\theta\right)}}, \mathrm{d}x=a\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \\&=&a \int \sqrt{a^2\left\{\tan^2{\left(\theta\right)}+1\right\}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \\&=&a \int a\sqrt{\tan^2{\left(\theta\right)}+1}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \\&=&a^2 \int \sqrt{\frac{1}{\cos^2{\left(\theta\right)}}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \;\cdots\;\tan^2{\left(\theta\right)}+1 =\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1 =\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} =\frac{1}{\cos^2{\left(\theta\right)}} \\&=&a^2 \int \frac{1}{\cos{\left(\theta\right)}}\;\frac{1}{\cos^2{\left(\theta\right)}}\mathrm{d}\theta \\&=&a^2 \int \frac{1}{\cos^3{\left(\theta\right)}}\mathrm{d}\theta \\&=&a^2 \left[\frac{1}{2}\left\{ \tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}} +\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|} \right\}+C_0 \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/cossec_25.html}{\int \frac{1}{\cos^3{\left(\theta\right)}}\mathrm{d}\theta =\frac{1}{2}\left\{ \tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}} +\ln{\left| \tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}} \right|} \right\}+C_0\;(C_0:積分定数)} \\&=&\frac{a^2}{2}\left\{ \tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}} +\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|} +C_0 \right\} \\&=&\frac{1}{2}\left\{ a^2\tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}} +a^2\ln{\left|\tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}\right|} +a^2C_0 \right\} \\&=&\frac{1}{2}\left\{ a^2\;\frac{1}{a^2}x\sqrt{x^2+a^2} +a^2\ln{\left| \frac{1}{a}\left(x+\sqrt{x^2+a^2}\right) \right|} +a^2C_0 \right\} \;\cdots\; \tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}=\frac{1}{a^2}x\sqrt{x^2+a^2}, \tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}}=\frac{1}{a}\left(x+\sqrt{x^2+a^2}\right) \\&=&\frac{1}{2}\left[ x\sqrt{x^2+a^2} +a^2\left\{ \ln{\left|x+\sqrt{x^2+a^2}\right|} -\ln{\left|a\right|} \right\} +a^2C_0 \right] \;\cdots\;\ln{\frac{A}{B}}=\ln{A}-\ln{B} \\&=&\frac{1}{2}\left\{ x\sqrt{x^2+a^2} +a^2\ln{\left|x+\sqrt{x^2+a^2}\right|} -a^2\ln{\left|a\right|} +a^2C_0 \right\} \\&=&\frac{1}{2}\left\{ x\sqrt{x^2+a^2} +a^2\ln{\left|x+\sqrt{x^2+a^2}\right|} \right\} -\frac{a^2}{2}\ln{\left|a\right|} +\frac{a^2}{2}C_0 \\&=&\frac{1}{2}\left\{ x\sqrt{x^2+a^2} +a^2\ln{\left|x+\sqrt{x^2+a^2}\right|} \right\} +C\;\cdots\;C=-\frac{a^2}{2}\ln{\left|a\right|}+\frac{a^2}{2}C_0\;(C:積分定数) \end{eqnarray} $$

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$$ \begin{eqnarray} \tan{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}} &=&\tan{\left(\theta\right)}\sqrt{\frac{1}{\cos^2{\left(\theta\right)}}} \\&=&\tan{\left(\theta\right)}\sqrt{\tan^2{\left(\theta\right)}+1} \;\cdots\; \frac{1}{\cos^2{\left(\theta\right)}} =\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} =\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1 =\tan^2{\left(\theta\right)}+1 \\&=&\frac{x}{a}\sqrt{\left(\frac{x}{a}\right)^2+1} \;\cdots\;x=a\tan{\left(\theta\right)}より\tan{\left(\theta\right)}=\frac{x}{a} \\&=&\frac{x}{a}\sqrt{\frac{x^2+a^2}{a^2}} \\&=&\frac{x}{a}\sqrt{\frac{1}{a^2}\left(x^2+a^2\right)} \\&=&\frac{x}{a}\frac{1}{a}\sqrt{x^2+a^2} \\&=&\frac{1}{a^2}x\sqrt{x^2+a^2} \end{eqnarray} $$

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$$ \begin{eqnarray} \tan{\left(\theta\right)}+\frac{1}{\cos{\left(\theta\right)}} &=&\tan{\left(\theta\right)}+\sqrt{\frac{1}{\cos^2{\left(\theta\right)}}} \\&=&\tan{\left(\theta\right)}+\sqrt{\tan^2{\left(\theta\right)}+1} \;\cdots\; \frac{1}{\cos^2{\left(\theta\right)}} =\frac{\sin^2{\left(\theta\right)}+\cos^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} =\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}+1 =\tan^2{\left(\theta\right)}+1 \\&=&\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2+1} \;\cdots\;x=a\tan{\left(\theta\right)}より\tan{\left(\theta\right)}=\frac{x}{a} \\&=&\frac{x}{a}+\sqrt{\frac{x^2+a^2}{a^2}} \\&=&\frac{x}{a}+\sqrt{\frac{1}{a^2}\left(x^2+a^2\right)} \\&=&\frac{x}{a}+\frac{1}{a}\sqrt{x^2+a^2} \\&=&\frac{1}{a}\left(x+\sqrt{x^2+a^2}\right) \end{eqnarray} $$

cosの逆数(sec)の三乗の積分

$$ \begin{eqnarray} \int \frac{1}{\cos^3{\left(\theta\right)}} \mathrm{d}\theta &=&\int \sec^3{\left(\theta\right)} \mathrm{d}\theta \;\cdots\;\sec{\left(\theta\right)}=\frac{1}{\cos{\left(\theta\right)}} \\&=&\int \frac{1}{\cos{\left(\theta\right)}}\;\frac{1}{\cos^2{\left(\theta\right)}} \mathrm{d}\theta \\&=&\int \frac{1}{\cos{\left(\theta\right)}}\;\left\{\tan{\left(\theta\right)}\right\}^\prime \mathrm{d}\theta \;\cdots\;\left\{\tan{\left(\theta\right)}\right\}^\prime =\left\{\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}\right\}^\prime =\href{https://shikitenkai.blogspot.com/2020/02/blog-post.html}{\left\{\sin{\left(\theta\right)}\right\}^\prime\frac{1}{\cos{\left(\theta\right)}}+\sin{\left(\theta\right)}\left\{\frac{1}{\cos{\left(\theta\right)}}\right\}^\prime} =\cos{\left(\theta\right)}\frac{1}{\cos{\left(\theta\right)}}+\sin{\left(\theta\right)}\frac{\sin{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} =\frac{\cos{\left(\theta\right)}}{\cos{\left(\theta\right)}}+\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} =\frac{\cos^{2}{\left(\theta\right)}+\sin^{2}{\left(\theta\right)}}{\cos^{2}{\left(\theta\right)}} =\frac{1}{\cos^{2}{\left(\theta\right)}} \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{\sin{\left(\theta\right)}}{\cos^2{\left(\theta\right)}}\;\tan{\left(\theta\right)} \mathrm{d}\theta \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post_7.html}{\int f^\prime g\;\mathrm{d}x=\left[fg\right]- \int f g^\prime \mathrm{d}x} \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{1}{\cos{\left(\theta\right)}}\tan^2{\left(\theta\right)} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{1}{\cos{\left(\theta\right)}}\;\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{\sin^2{\left(\theta\right)}}{\cos^3{\left(\theta\right)}} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{1-\cos^2{\left(\theta\right)}}{\cos^3{\left(\theta\right)}} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \left\{ \frac{1}{\cos^3{\left(\theta\right)}}-\frac{\cos^2{\left(\theta\right)}}{\cos^3{\left(\theta\right)}} \right\} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \left\{ \frac{1}{\cos^3{\left(\theta\right)}}-\frac{1}{\cos{\left(\theta\right)}} \right\} \mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} -\int \frac{1}{\cos^3{\left(\theta\right)}} \mathrm{d}\theta +\int \frac{1}{\cos{\left(\theta\right)}} \mathrm{d}\theta \\ 2\int \frac{1}{\cos^3{\left(\theta\right)}} \mathrm{d}\theta &=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} +\int \frac{1}{\cos{\left(\theta\right)}}\mathrm{d}\theta \\&=&\frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} +\frac{1}{2}\ln{\left| \frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right|} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/cossec.html}{\int \frac{1}{\cos{\left(\theta\right)}}\mathrm{d}\theta =\frac{1}{2}\ln{\left| \frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right|}+C\;(C:積分定数) } \\ \int \frac{1}{\cos^3{\left(\theta\right)}}\mathrm{d}\theta &=&\frac{1}{2}\left\{ \frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} +\frac{1}{2}\ln{\left| \frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right|} \right\}+C\;\cdots\;C:積分定数 \\&=&\frac{1}{2}\left\{ \frac{\tan{\left(\theta\right)}}{\cos{\left(\theta\right)}} +\ln{\left| \frac{1}{\cos{\left(\theta\right)}}+\tan{\left(\theta\right)} \right|} \right\}+C \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/cossec.html}{\frac{ 1+\sin{\left(\theta\right)} }{ 1-\sin{\left(\theta\right)} }=\left\{ \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right\}^2 } \\&=&\frac{1}{2}\left\{ \sec{\left(\theta\right)}\tan{\left(\theta\right)} +\ln{\left| \sec{\left(\theta\right)}+\tan{\left(\theta\right)} \right|} \right\}+C \;\cdots\;\sec{\left(\theta\right)}=\frac{1}{\cos{\left(\theta\right)}} \end{eqnarray} $$

cosの逆数(sec)の積分

$$ \begin{eqnarray} \int \frac{1}{\cos{\left(\theta\right)}} \mathrm{d}\theta &=&\int \sec{\left(\theta\right)} \mathrm{d}\theta \;\cdots\;\sec{\left(\theta\right)}=\frac{1}{\cos{\left(\theta\right)}} \\&=&\int \frac{1}{\cos{\left(\theta\right)}} \frac{\cos{\left(\theta\right)}}{\cos{\left(\theta\right)}} \mathrm{d}\theta \\&=& \int \frac{\cos{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} \mathrm{d}\theta \\&=& \int \frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}} \mathrm{d}\theta \;\cdots\;\cos^2{\left(\theta\right)}=1-\sin^2{\left(\theta\right)} \\&=& \int \frac{1}{2} \left\{ \frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}} +\frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right\} \mathrm{d}\theta \\&&\;\cdots\;\frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}} =\frac{\alpha}{1+\sin{\left(\theta\right)}} +\frac{\beta}{1-\sin{\left(\theta\right)}} \\&&\;\cdots\;\cos{\left(\theta\right)} =\alpha\left(1-\sin{\left(\theta\right)}\right)+\beta\left(1+\sin{\left(\theta\right)}\right) =\left(\alpha+\beta\right)+\left(\alpha-\beta\right)\sin{\left(\theta\right)} \\&&\;\cdots\; \left\{ \begin{array}{l} \cos{\left(\theta\right)}=\alpha+\beta \\0=\left(\alpha-\beta\right)\sin{\left(\theta\right)} \end{array} \right. \\&&\;\cdots\;\sin{\left(\theta\right)}は\thetaによっては0とはならないので,\alpha-\betaが常に0,よって\alpha=\beta. \\&&\;\cdots\;従って\alpha+\beta=\alpha+\alpha=2\alpha=\cos{\left(\theta\right)}. \\&&\;\cdots\;\alpha=\beta=\frac{1}{2}\cos{\left(\theta\right)} \\&&\;\cdots\;\frac{\cos{\left(\theta\right)}}{1-\sin^2{\left(\theta\right)}} =\frac{\frac{1}{2}\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}} +\frac{\frac{1}{2}\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \\&=& \frac{1}{2} \int \left\{ \frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}} +\frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right\} \mathrm{d}\theta \\&=& \frac{1}{2} \left\{ \int \frac{\cos{\left(\theta\right)}}{1+\sin{\left(\theta\right)}}\mathrm{d}\theta +\int \frac{\cos{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}\mathrm{d}\theta \right\} \\&=& \frac{1}{2} \left\{ \int \frac{\cos{\left(\theta\right)}}{u}\frac{1}{\cos{\left(\theta\right)}}\mathrm{d}u +\int \frac{\cos{\left(\theta\right)}}{v}\frac{-1}{\cos{\left(\theta\right)}}\mathrm{d}v \right\} \\&&\;\cdots\;u=1+\sin{\left(\theta\right)},\frac{\mathrm{d}u}{\mathrm{d}\theta}=\cos{\left(\theta\right)},\mathrm{d}\theta=\frac{1}{\cos{\left(\theta\right)}}\mathrm{d}u \\&&\;\cdots\;v=1-\sin{\left(\theta\right)},\frac{\mathrm{d}v}{\mathrm{d}\theta}=-\cos{\left(\theta\right)},\mathrm{d}\theta=\frac{-1}{\cos{\left(\theta\right)}}\mathrm{d}v \\&=& \frac{1}{2} \left( \int \frac{1}{u} \mathrm{d}u - \int \frac{1}{v} \mathrm{d}v \right) \\&=& \frac{1}{2} \left( \ln{\left|u\right|} - \ln{\left|v\right|} \right)\;\cdots\;\int \frac{1}{x} \mathrm{d}x=\ln{\left|x\right|}+C\;(C:積分定数) \\&=& \frac{1}{2} \ln{\frac{\left|u\right|}{\left|v\right|}}\;\cdots\;\log{A}-\log{B}=\log{\frac{A}{B}} \\&=& \frac{1}{2} \ln{\left|\frac{u}{v}\right|}\;\cdots\;\frac{\left|A\right|}{\left|B\right|}=\left|\frac{A}{B}\right| \\&=& \frac{1}{2} \ln{\left|\frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}}\right|} + C \;\cdots\;C:積分定数 \\もしくは， \\&=& \frac{1}{2} \ln{\left| \frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \right|}+ C \;\cdots\;A\log{B}=\log{B^A} \\&=& \frac{1}{2} \ln{\left| \left\{ \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right\}^2 \right|}+C \;\cdots\;\frac{ 1+\sin{\left(\theta\right)} }{ 1-\sin{\left(\theta\right)} }=\left\{ \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right\}^2 \;式変形は最後に記載 \\&=& \frac{1}{2} 2 \ln{\left| \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right|}+C \;\cdots\;\left(A^B\right)^\frac{1}{B}=A^\left(B\frac{1}{B}\right)=A^1=A \\&=& \ln{\left| \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right|}+C \\&=& \ln{\left| \sec{\left(\theta\right)} +\tan{\left(\theta\right)} \right|}+C \;\cdots\;\frac{1}{\cos{\left(\theta\right)}}=\sec{\left(\theta\right)} \end{eqnarray} $$

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$$ \begin{eqnarray} \frac{ 1+\sin{\left(\theta\right)} }{ 1-\sin{\left(\theta\right)} } &=& \frac{1+\sin{\left(\theta\right)}}{1-\sin{\left(\theta\right)}} \;\frac{1+\sin{\left(\theta\right)}}{1+\sin{\left(\theta\right)}} \\&=& \frac{ \left\{1+\sin{\left(\theta\right)}\right\}^2 }{ 1-\sin^2{\left(\theta\right)} } \;\cdots\;(A+B)(A-B)=A^2-B^2 \\&=& \frac{ 1+2\sin{\left(\theta\right)}+\sin^2{\left(\theta\right)} }{ \cos^2{\left(\theta\right)} } \;\cdots\;1-\sin^2{\left(\theta\right)}=\cos^2{\left(\theta\right)},\;(A+B)^2=A^2+2AB+B^2 \\&=& \frac{1}{\cos^2{\left(\theta\right)}} +2\frac{\sin{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} +\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} \\&=& \frac{1}{\cos^2{\left(\theta\right)}} +2\frac{1}{\cos{\left(\theta\right)}}\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}} +\frac{\sin^2{\left(\theta\right)}}{\cos^2{\left(\theta\right)}} \\&=& \frac{1}{\cos^2{\left(\theta\right)}} +2\frac{1}{\cos{\left(\theta\right)}}\tan{\left(\theta\right)} +\tan^2{\left(\theta\right)} \;\cdots\;\frac{\sin{\left(\theta\right)}}{\cos{\left(\theta\right)}}=\tan{\left(\theta\right)} \\&=& \left\{ \frac{1}{\cos{\left(\theta\right)}} +\tan{\left(\theta\right)} \right\}^2 \;\cdots\;A^2+2AB+B^2=(A+B)^2 \end{eqnarray} $$

多角形の頂点座標から重心を求める (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。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

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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． $$ \begin{eqnarray} Y_G&=&-\frac{1}{2A}\oint_C y^2\mathrm{d}x \\&=&-\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} y^2 \mathrm{d}x \end{eqnarray} $$

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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． $$ \begin{eqnarray} y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right) \;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p) \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \end{eqnarray} $$ $$ \begin{eqnarray} \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1} \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}{\color{red}+x_{k-1}y_{k-1}}+x_{k-1}y_k{\color{red}-x_{k-1}y_{k-1}}}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}} \end{eqnarray} $$ $$ \begin{eqnarray} y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right) \\y&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)+y_{k-1} \\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{y_k-y_{k-1}}{x_k-x_{k-1}}(-x_{k-1})+y_{k-1} \\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{(-x_{k-1})(y_k-y_{k-1})+(x_k-x_{k-1})y_{k-1}}{x_k-x_{k-1}} \\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}y_k{\color{red}+x_{k-1}y_{k-1}}+x_ky_{k-1}{\color{red}-x_{k-1}y_{k-1}}}{x_k-x_{k-1}} \\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{-x_{k-1}y_k+x_ky_{k-1}}{x_k-x_{k-1}} \\&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}} \end{eqnarray} $$

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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． $$ \begin{eqnarray} Y_G&=&-\frac{1}{2A}\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} y^2 \mathrm{d}y \\&=&-\frac{1}{2A}\sum_{k=1}^n \int^{x_k}_{x_{k-1}} \left( \frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}} \right)^2 \mathrm{d}y \;\cdots\;y=\frac{y_k-y_{k-1}}{x_k-x_{k-1}}x+\frac{x_ky_{k-1}-x_{k-1}y_k}{x_k-x_{k-1}} \\&=&-\frac{1}{6A}\sum_{k=1}^n \left(y_k^2 +y_ky_{k-1} +y_{k-1}^2\right) \left(x_k -x_{k-1}\right) \end{eqnarray} $$

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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． $$ \begin{eqnarray} (X_G,Y_G)&=&(\frac{1}{2},\frac{1}{2}) \end{eqnarray} $$

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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． $$ \begin{eqnarray} (X_G,Y_G)&=&(\frac{1}{2},\frac{1}{3}) \end{eqnarray} $$

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。 $$ \begin{eqnarray} \iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y &=& \oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y \end{eqnarray} $$

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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 $$ \begin{eqnarray} A&=&\iint_D 1\;\mathrm{d}x\mathrm{d}y \end{eqnarray} $$ 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． $$ \begin{eqnarray} A&=&\oint_C x\mathrm{d}y \\&=&\sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} x \mathrm{d}y \end{eqnarray} $$

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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． $$ \begin{eqnarray} y_-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right) \;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p) \\\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)&=&x-x_{k-1} \\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right) \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1} \\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}\color{red}{+x_{k-1}y_{k-1}} \color{black}{+x_{k-1}y_k} \color{red}{-x_{k-1}y_{k-1}} }{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}} \\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}} \end{eqnarray} $$

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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{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．

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例:正四角形

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

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例:二等辺三角形

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