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Parametric and Polar Curves

The Area Below a Polar Curve

Problems

Problems

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)    

Figure %: Point with Polar Coordinates [r, θ]

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).

Figure %: Finding the Area of a Polar Curve

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    

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 = θ =    

which is indeed one quarter of the area of a unit circle.

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