Answer to Question #98856 in Calculus for Rachel

Question

In this question, we are going to derive the Surveyor’s Formula for finding the area of a polygon in terms of its coordinates.(a) If C is the line segment from (x1, y1) to (x2, y2) and ω=−ydx+xdy is a differential 1-form, show that ∫Cω=x1y2−x2y1 (b) If the vertices of a polygon (in counterclockwise order) are (x1, y1),(x2, y2), . . .(xn, yn),show that the area of the polygon is A=12[(x1y2−x2y1) + (x2y3−x3y2) +. . .(xn−1yn−xnyn−1) + (xny1+x1yn)] (c) Find the area of a pentagon with vertices (0,0),(2,1),(1,3),(0,2),(−1,1).

Answer 1

a.Parametrize the straight line C by$x = x_1 + t(x_2 − x_1)$ ,

$y = y_1 + t(y_2 − y_1),$

where t goes from $t = 0$ to $t = 1$ . Therefore

$\oint ydx+xdy=\int_{0} ^{1} −(y_1 + t(y_2 − y_1))(x_2 − x_1)dt +\int_{0} ^{1} (x_1 + t(x_2 − x_1))(y_2 − y_1)dt$

$= −y_1(x_2 − x_1) −(y_2 − y_1)(x_2 − x_1)/2+ x1(y_2 − y_1) + (x_2 − x_1)(y_2 − y_1)$

$= −y_1(x_2 − x_1) + x_1(y_2 − y_1) = x_1y_2 − x_2y_1$ (Proved)

b.Consider two consecutive vertices$(x_i, y_i)$ and$(x_{i+1}, y_{i+1})$ . Let C_i be the line segment joining $(x_i,y_i)$ to $(x_{i+1}, y_{i+1})$ $, i = 1, . . . , n,$ with the convention that$(x_{n+1}, y_{n+1}) = (x_1, y_1)$ . Let us compute

$\oint (xdy-ydx)$

To do so, we can parametrize $C_i$ as

$x_i=(1-t)x_i + tx_{i+1}$

$y_i= (1 − t)y_i + ty_{i+1},$

$dx = (x_{i+1} − x_i) dt$

$dy = (y_{i+1 }− y_i) dt$

$\oint (xdy-ydx)= \int_{0} ^{1} ((1-t)x_i + tx_{i+1}).(y_{i+1 }− y_i) dt- \int_{0} ^{1} ((1 − t)y_i + ty_{i+1}). (x_{i+1} − x_i) dt$

$=\int_{0} ^{1}x_iy_{i+1}-x_i+2y_i$

$=x_iy_{i+1}-x_{i+1}$ -----(1)

Denoting by D the region enclosed by the polygon

We know $A=1/2\smallint _{dD} ( xdy-ydx)$

$=1/2\Sigma _{i=1}^n (\smallint _{C_i}(xdy-ydx))$

Using (1): we get the desired result

$A=1/2[(x_1y_2−x_2y_1) + (x_2y_3−x_3y_2) +. . .(x_{n−1}y_n−x_ny_{n−1}) + (x_ny_1+x_1y_n)]$

(Proved)

(c)$Area=1/2[(0-0)+(2.3-1.1)+(1.2-0)+(0-(-2))+(0-0)]$

$=1/2(5+2+2)=9/2$ (Answer) using formula proved in (b)