The graph of r=1 intersects the graph of r=2sin(theta) when
2sin(theta)=1
sin(theta)=1/2
theta = 2pi n + pi/6 for n in ZZ
Looking at the graph, we can see that those angles correspond the intersection in quadrant I and quadrant II.
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Analyzing the graph, we can develop a strategy for finding the area of the overlapping regions. For example, we could get the area of r=1 (in red) from theta=pi//6 to theta=5pi//6. Then we would need to add on both remaining slivers of r = 2sin(theta) (in blue). Those two slivers are identical in area, so we could add the area of r = 2sin(theta) between theta=0 and theta=pi//6, then double it.
This gives us
A=1/2int_(pi/6)^((5pi)/6)(1)^2d theta + 2 *1/2int_0^(pi/6) (2sin(theta))^2 d theta
A=[1/2 theta]_(pi/6)^((5pi)/6)-4[1/2 (θ - sin(θ) cos(θ))]_0^(pi/6)
A=(4pi)/12-4[1/24 (2 π - 3 sqrt(3))]~~sqrt(3)/2
