How can you calculate the area of a polygon given its vertices?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$ be points in $\mathbb{R}^2$ such that when they are joined together in succession by line segments [and $(x_n,y_n)$ is joined to $(x_1,y_1)$], the segments enclose a polygonal region $R$ (see attached picture).

View attachment 500​

Assuming that the polygonal boundary is counterclockwise oriented, prove that the area of the region $R$ is given in terms of the coordinates of its vertices as follows:

\[A(R)=\frac{1}{2}\left(\begin{vmatrix}x_1 & x_2\\ y_1 & y_2\end{vmatrix}+\begin{vmatrix}x_2 & x_3\\ y_2 & y_3\end{vmatrix}+\cdots+\begin{vmatrix}x_{n-1} & x_n\\ y_{n-1} & y_n\end{vmatrix}+\begin{vmatrix}x_n & x_1\\ y_n & y_1\end{vmatrix}\right)\]

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Hint:

Spoiler

Use Green's Theorem.

 

[Chris L T521]

P.S: For you curious ones out there, here's the TeX code I used to make that image. [I originally was going to put this in the original question post, but it exceeded the character limit... >_>]

\begin{figure}
   \centering
   \begin{tikzpicture}[domain=-6:6]
      \draw[->] (-3,-1) -- (-3,5) node[above]{$y$};
      \draw[->] (-4,0) -- (5.5,0) node[right]{$x$};
      \fill (2,.5) -- (3.5,1.5) -- (4,3) -- (2.5,4) -- (0,4) -- (-1.5,2.5) -- (-1, 1.5) -- (0,2) -- (1.5,1.5);
      \draw (2,.5) -- (3.5,1.5) -- (4,3) -- (2.5,4) -- (0,4) -- (-1.5,2.5) -- (-1, 1.5) -- (0,2) -- (1.5,1.5) -- (2,.5);
      \fill (2,.5) circle (2pt);
      \fill (3.5,1.5) circle (2pt);
      \fill (4,3) circle (2pt);
      \fill (2.5,4) circle (2pt);
      \fill (0,4) circle (2pt);
      \fill (-1.5,2.5) circle (2pt) node[right]{$\color{black}(x_n,y_n)$};
      \fill (-1, 1.5) circle (2pt) node[below]{$\color{black}(x_1,y_1)$};
      \fill (0,2) circle (2pt)  node[above=.25cm,right]{$\color{black}(x_2,y_2)$};
      \fill (1.5,1.5) circle (2pt) node[right]{$\color{black}(x_3,y_3)$};
      \draw (2,.5) -- (2.75,1);
      \draw (3.5,1.5) -- (3.75,2.25);
      \draw (4,3) -- (3.25,3.5);
      \draw (2.5,4) -- (1.25,4);
      \draw (0,4) -- (-.75,3.25);
      \draw (-1.5,2.5) -- (-1.25,2);
      \draw (-1,1.5) -- (-.5,1.75);
      \draw (0,2) -- (.75,1.75);
      \draw (1.5,1.5) -- (1.75,1);
   \end{tikzpicture}
   
\end{figure}

To get it to compile, you'll need the tikz package!)

 

[Chris L T521]

This week's question was correctly answered by MarkFL and Sudharaka.

Here's MarkFL's solution:

Spoiler

Using Green's theorem, we may state:

$\displaystyle A(R)=\int_{R}\langle 0,x \rangle\cdot dr=\int_{R}\langle -y,0 \rangle\cdot dr$

Let $\displaystyle R_k$ be an arbitrary side of $\displaystyle R$, which we may parametrize as follows:

$\displaystyle r_k(t)=\langle x_k+t(x_{k+1}-x_k),y_k+t(y_{k+1}-y_k) \rangle$ where $\displaystyle 0\le t\le1$

where $\displaystyle 1\le k\le n$ and $\displaystyle (x_{n+1},y_{n+1})=(x_1,y_1)$.

Using the two integrals above, we may state:

(1) $\displaystyle A(R)=\int_{R}\langle 0,x \rangle\cdot dr=\sum_{k=1}^n\left(\int_{R_k}\langle 0,x \rangle\cdot dr \right)$

(2) $\displaystyle A(R)=\int_{R}\langle -y,0 \rangle\cdot dr=\sum_{k=1}^n\left(\int_{R_k}\langle -y,0 \rangle\cdot dr \right)$

Using:

$\displaystyle r_k'(t)=\langle x_{k+1}-x_k,y_{k+1}-y_k \rangle$

we now have:

$\displaystyle \int_{R_k}\langle 0,x \rangle\cdot dr=\int_0^1 \langle 0,x_k+t(x_{k+1}-x_k) \rangle\cdot\langle x_{k+1}-x_k,y_{k+1}-y_k \rangle\,dt$

$\displaystyle \int_{R_k}\langle 0,x \rangle\cdot dr=(y_{k+1}-y_k)\int_0^1 x_k+t(x_{k+1}-x_k)\,dt$

$\displaystyle \int_{R_k}\langle 0,x \rangle\cdot dr=(y_{k+1}-y_k)\left[x_kt+(x_{k+1}-x_k)\frac{t^2}{2} \right]_0^1$

$\displaystyle \int_{R_k}\langle 0,x \rangle\cdot dr=(y_{k+1}-y_k)\left(x_k+\frac{1}{2}(x_{k+1}-x_k) \right)$

$\displaystyle \int_{R_k}\langle 0,x \rangle\cdot dr=\frac{1}{2}(y_{k+1}-y_k)(x_k+x_{k+1})$

Using our summation formula (1), we find:

$\displaystyle A(R)=\sum_{k=1}^n\left(\frac{1}{2}(y_{k+1}-y_k)(x_k+x_{k+1}) \right)$

Using the second integral, we have:

$\displaystyle \int_{R_k}\langle -y,0 \rangle\cdot dr=\int_0^1 \langle -(y_k+t(y_{k+1}-y_k)),0 \rangle\cdot\langle x_{k+1}-x_k,y_{k+1}-y_k \rangle\,dt$

$\displaystyle \int_{R_k}\langle -y,0 \rangle\cdot dr=(x_{k+1}-x_k)\int_0^1 -(y_k+t(y_{k+1}-y_k)\,dt$

$\displaystyle \int_{R_k}\langle -y,0 \rangle\cdot dr=-(x_{k+1}-x_k)\left[y_kt+(y_{k+1}-y_k)\frac{t^2}{2} \right]_0^1$

$\displaystyle \int_{R_k}\langle -y,0 \rangle\cdot dr=-(x_{k+1}-x_k)\left(y_k+\frac{1}{2}(y_{k+1}-y_k) \right)$

$\displaystyle \int_{R_k}\langle -y,0 \rangle\cdot dr=\frac{1}{2}(x_k-x_{k+1})(y_k+y_{k+1})$

Using our summation formula (2), we find:

$\displaystyle A(R)=\sum_{k=1}^n\left(\frac{1}{2}(x_k-x_{k+1})(y_k+y_{k+1}) \right)$

Adding the two summations, there results:

$\displaystyle 2A(R)=\sum_{k=1}^n\left(\frac{1}{2}(y_{k+1}-y_k)(x_k+x_{k+1}) \right)+\sum_{k=1}^n\left(\frac{1}{2}(x_k-x_{k+1})(y_k+y_{k+1}) \right)$

$\displaystyle 4A(R)=\sum_{k=1}^n\left((y_{k+1}-y_k)(x_k+x_{k+1})+(x_k-x_{k+1})(y_k+y_{k+1}) \right)$

$\displaystyle 4A(R)=2\sum_{k=1}^n\left(x_ky_{k+1}-x_{k+1}y_k \right)$

$\displaystyle 2A(R)=\sum_{k=1}^n\left(\begin{vmatrix}x_k & x_{k+1}\\ y_k & y_{k+1}\end{vmatrix} \right)$

Hence:

$\displaystyle A(R)=\frac{1}{2}\left(\begin{vmatrix}x_1 & x_2\\ y_1 & y_2\end{vmatrix}+\begin{vmatrix}x_2 & x_3\\ y_2 & y_3\end{vmatrix}+\cdots+\begin{vmatrix}x_{n-1} & x_n\\ y_{n-1} & y_n\end{vmatrix}+\begin{vmatrix}x_n & x_1\\ y_n & y_1\end{vmatrix}\right)$

Shown as desired.

Here's Sudharaka's solution:

Spoiler

Using the Green's theorem with \(M=\dfrac{x}{2}\) and \(L=-\dfrac{y}{2}\) we get,

\[A=\iint_{R}\mathrm{d}x\, \mathrm{d}y=\frac{1}{2}\oint_{C} (-y\, \mathrm{d}x + x\, \mathrm{d}y)\]

where \(C\) is the boundary of the region \(R\).

\[\therefore A=\frac{1}{2}\sum_{k=1}^{n}\int_{A_1}^{A_2}(-y\, \mathrm{d}x + x\, \mathrm{d}y)\]

where \(A_{1}=(x_{k},\, y_{k})\) and \(A_2=(x_{k+1},\, y_{k+1})\) with \((x_{n+1},\,y_{n+1})=(x_{1},\,y_{1})\).

The equation of the line joining the points, \(A_1\) and \(A_2\) is,

\[y=y_{k+1}+\left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)(x-x_{k+1})\]

\[\therefore dy=\left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)dx\]

Hence we get,

\begin{eqnarray}

A&=&\frac{1}{2}\sum_{k=1}^{n}\int_{A_1}^{A_2} \left[-y_{k+1}-\left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)(x-x_{k+1})+x \left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)\right]\, dx\\

&=&\frac{1}{2}\sum_{k=1}^{n}\int_{A_1}^{A_2} \left[-y_{k+1}+\left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)x_{k+1}\right]\, dx\\

&=&\frac{1}{2}\sum_{k=1}^{n}\left\{\left[-y_{k+1}+\left(\frac{y_{k+1}-y_{k}}{x_{k+1}-x_{k}}\right)x_{k+1}\right](x_{k+1}-x_{k})\right\}\\

&=&\frac{1}{2}\sum_{k=1}^{n}(y_{k+1}x_k-y_{k}x_{k+1})\\

&=&\frac{1}{2}\sum_{k=1}^{n}\begin{vmatrix}x_{k} & x_{k+1}\\ y_{k} & y_{k+1}\end{vmatrix}\\

&=&\frac{1}{2}\left(\begin{vmatrix}x_1 & x_2\\ y_1 & y_2\end{vmatrix}+\begin{vmatrix}x_2 & x_3\\ y_2 & y_3\end{vmatrix}+\cdots+\begin{vmatrix}x_{n-1} & x_n\\ y_{n-1} & y_n\end{vmatrix}+\begin{vmatrix}x_n & x_1\\ y_n & y_1\end{vmatrix}\right)

\end{eqnarray}