Polar coordinates provide an alternate way of specifying a point in the plane. The polar coordinates [r, θ] represent the point at a distance r from the origin, rotated θ radians counterclockwise from the positive x -axis. Since r represents a distance, it is typically positive. Sometimes, r is allowed to be negative; in this case [r, θ] represents the reflection about the origin of the point [| r|, θ] .
It follows from basic trigonometry that the point with polar coordinates [r, θ] has Cartesian coordinates
(r cosθ, r sinθ) |
Going the other direction, the point with Cartesian coordinates (x, y) has polar coordinates
, tan^{-1} |
if it lies in quadrants I or IV and polar coordinates
, tan^{-1} + Π |
if it lies in quadrants II or III .
A polar function r(θ) has a graph consisting of the points [r(θ), θ] . Such a graph is known as a polar curve. One of the simplest polar curves is the circle, the graph of the polar function r(θ) = c , for some constant c . In the remainder of this section, we investigate how to find the area enclosed by a polar curve from one value of θ to another. For example, we might wish to find the area of the region below the curve r(θ) = 1 from θ = 0 to θ = Π/2 (this region is of course a quarter of the interior of a unit circle).
Considering the general case, the idea is similar to the idea for finding the area below the graph of a function in Cartesian coordinates. In that case, we approximated the region by a bunch of thin rectangles; here, we approximate it by thin circular sectors (shaped like slices of pie).
Such a method worked before because we knew beforehand how to compute the area of a rectangle. Now we attempt this computation for a circular sector. Suppose the sector has angular width of Δθ and is part of a circle of radius r , with area Πr ^{2} . Since the sector accounts for Δθ/2Π of the area of the circle, the area of the sector is equal to
Πr ^{2} = (Δθ)r ^{2} |
Summing together the areas of all the thin sectors and taking the limit as Δθ→ 0 (and the number of sectors approaches infinity), we get the definite integral
r(θ)^{2} dθ |
Note that, because of the square in the expression being integrated, the integral counts all area as positive, even when r(θ) < 0 .
Applying this theory to the example given above, we get an area of
(1)^{2} dθ = θ = |
which is indeed one quarter of the area of a unit circle.