264 CHAPTER 12. POLAR COORDINATES

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

12

sin(dθ) f (θ)2 ≈ 12

f (θ)2 dθ = dA

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

dAdθ

=12

f (θ)2 , A(a) = 0.

Therefore, the total area would be given by the integral

12

∫ b

af (θ)2 dθ . (12.2)

Example 12.2.1 Find the area of the cardioid, r = 1+ cosθ for θ ∈ [0,2π].

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

12

∫ 2π

0(1+ cos(θ))2 dθ =

32

π.

Example 12.2.2 Verify the area of a circle of radius a is πa2.

The polar equation is just r = a for θ ∈ [0,2π]. Therefore, the area should be

12

∫ 2π

0a2dθ = πa2.

Example 12.2.3 Find the area of the region inside the cardioid, r = 1+ cosθ and outsidethe circle r = 1 for θ ∈

[−π

2 ,π

2

].

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

-1 0 1 2

-1

0

1

desiredregion

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

[−π

2 ,π

2

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

12

∫π/2

−π/2(1+ cos(θ))2 dθ − 1

2

∫π/2

−π/21dθ =

14

π +2.

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