$title="Oriented Area Part 1"; $page_name="oriented1"; include "header.php"; ?>
Most commonly used geometry technology can easily handle the area of general polygons, given by a sequence of points in the plane, without ever considering convexity or self-intersection. The way they do this is by using formulas that assign to each triangle an "algebraic area," which is the usual area if the triangle is oriented in a counterclockwise matter, and the negative of that area if the triangle is oriented clockwise. It is worth noting that this choice of sign is completely arbitrary: things would work just as well if the opposite convention were chosen.figure_indent("images/oriented_area.jpg", "", "Figure", "1", "Triangles of opposite orientation have areas of opposite sign."); ?>
Polygons with more than three sides can be broken up into triangles, the sum of whose areas is the area of the polygon itself. For quadrilaterals, we may simply choose one of the diagonals and calculate the areas of the two triangles thus formed. Either choice of diagonal will yield the same result. Here is a table illustrating how three types of quadrilaterals (convex, non-convex, and self-intersecting) can be subdivided into triangles:figure_indent("images/oa_quad_sub.jpg", "", "Figure", "2", "Subdivision of quadrilaterals by choosing a diagonal. Note the orientation of each triangle."); ?>
This method for quadrilaterals is actually a special case of a method for n-gons: choose any point as the origin and construct a triangle to this point from each edge of the polygon. The sum of the oriented areas of these triangles is the oriented area of the polygon, and is independent of the choice of the origin. This method is illustrated below:figure_indent("images/oa_poly_sub.jpg", "", "Figure", "3", "General method for calculating the algebraic area of any polygon: choose a point A and construct triangles. The sum of the areas of the triangles is independent of the choice of A. The two triangular subdivisions of this hexagon are equivalent, since the blue shaded area on the left is counted once positively and once negatively."); ?>
A great advantage of this method is that it does not make any difference how we divide a given polygon up into a number of triangles. The sum of the algebraic areas of those triangles will always give the sum of the areas of the parts traversed counterclockwise minus the absolute areas of the parts traversed clockwise.
This fundamental intuition is at the heart of the computation of so-called "line integrals" which students will encounter in their third semester of calculus. When we teach this subject, often this is the first time that students are exposed to the concepts of oriented area for smooth curves, without an appreciation of the fact that the same concept works for polygons.
There are formulas for the area of a triangle in terms of the coordinates of the vertices that automatically give the algebraic area. For example, to find the algebraic area of a triangle with first vertex at (0, 0), second vertex at (a, b) and third vertex at (c, d), we may use the formula (1/2)(ad - bc). This will be positive if the slope d/c of the line from (0, 0) through (c, d) is greater than the slope b/a of the line from (0, 0) through (a, b), since d/c > b/a implies that ad > bc and ad - bc > 0. In terms of vectors, if the line along (a, b) rotates in a counterclockwise direction to get to the line along (c, d), then the triangle from (0, 0) to (a, b) to (c, d) has positive area, and if it rotates in a clockwise direction, the algebraic area of the triangle is negative.figure_indent("images/oa_tri_formula.jpg", "", "Figure", "4", "The area of a triangle with vertices (0, 0), (a, b), and (b, c) is given by (1/2)(ad - bc). Note that the counterclockwise-oriented triangle on the left has positive area, while the clockwise-oriented triangle has negative area."); ?>
For those familiar with vectors in physics, another way of expressing this algebraic area is to think of the three vertices as lying in a plane in ordinary three-dimensional space, and computing the cross-product between the vectors (a, b, 0) and (c, d, 0) to get (0, 0, ad-bc). The algebraic area is then obtained by taking half the dot product of this vector with (0, 0, 1). If the rotation from (a, b, 0) to (c, d, 0) is counterclockwise, then the cross-product points upward and the dot product with (0, 0, 1) is positive. If the rotation is negative, then the cross-product points in the direction opposite to (0, 0, 1) and the algebraic area is negative.
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