Area between two polar curves ( dont understand why this is the answer)

Question

Consider the curves:

 

γ1 : r(θ) = 1 + sinθ , 0 ≤ θ ≤2pi

 

γ2 : r(θ) = 3sinθ , 0 ≤ θ ≤2pi

 

Calculate the area in the first quadrant enclosed by the both curves and the y-axis. 

 

 

In my solutions booklet, it shows this as the answer : 

 

∫ 1/2(3sinθ)² dθ + ∫ 1/2(1+sinθ)² dθ

 

What I dont understand is how do I know which equation to put first . In this problem its the 3sinθ being squared first and 1+sinθ being the second squared. How do I determine the order 

Answer 1

Area = ½ ∫r²dθ

 

First, draw the region.

 

The curves intersect when 1 + sinθ = 3sinθ

 

                                            2sinθ = 1

 

                                              sinθ = 1/2    θ = π/6

 

Between 0 and π/6, the boundary is r = 3sinθ.

 

Between π/6 and π/2, the boundary is r = 1 + sinθ

 

Area enclosed by both curves in quadrant I:

 

   ½∫_{(from 0 to π/6)}(3sinθ)²dθ + ½∫_{(from π/6 to π/2)} (1+sinθ)²dθ