Calculus of One and Several Variables

12.2 The Area In Polar Coordinates

How can you find the area of the region determined by 0 ≤ r ≤ f

for θ ∈, assuming this is a well defined set of points in the plane? See Example 12.1.3 with θ ∈ to see something which it would be better to avoid.

I have in mind the situation where every ray through the origin having angle θ for θ ∈

,b−a ≤ 2π, intersects the graph of r = f in exactly one point. To see how to find the area of such a region, consider the following picture.

This is a representation of a small triangle obtained from two rays whose angles differ by only dθ. What is the area of this triangle, dA? It would be

1 sin(dθ)f (θ)2 ≈ 1 f (θ)2dθ = dA 2 2

with the approximation getting better as the angle gets smaller. Thus the area should solve the initial value problem,

dA-= 1 f (θ)2 ,A (a) = 0. dθ 2

Therefore, the total area would be given by the integral

∫ b 1 f (θ)2dθ. (12.2) 2 a

(12.2)

From the graph of the cardioid presented earlier, you can see the region of interest satisfies the conditions above that every ray intersects the graph in only one point. Therefore, from 12.2 this area is

∫ 2π 1 (1 + cos (θ))2dθ = 3π. 2 0 2

The polar equation is just r = a for θ ∈

. Therefore, the area should be

∫ 1 2π 2 2 2 0 a dθ = πa .

As is usual in such cases, it is a good idea to graph the curves involved to get an idea what is wanted.

The area of this region would be the area of the part of the cardioid corresponding to θ ∈

minus the area of the part of the circle in the first quadrant. Thus the area is

∫ π∕2 ∫ π∕2 1 (1 + cos(θ))2dθ− 1 1dθ = 1π+ 2. 2 − π∕2 2 −π∕2 4

This example illustrates the following procedure for finding the area between the graphs of two curves given in polar coordinates.