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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.