Find the area of the region that is inside the graph of r = 5 sin θ and outside of the graph of r = 2 + sin θ

Question

Find the area of the region that is inside the graph of r = 5 sin θ and outside of the graph of r = 2 + sin θ

A	10.109
B	3.323
C	6.646
D	20.219

Answer 1

Intersection: 5sinθ = 2 + sinθ; sinθ = 1/2, θ = π/6 or θ = 5π/6;

A = 1/2∫_π/6^5π/6[25sin²θ - (2 + sinθ)²]dθ = 1/2∫_π/6^5π/6(24sin²θ - 4 - 4sinθ)dθ =

= 1/2∫_π/6^5π/6(8 - 12cos2θ - 4sinθ)dθ = (4θ - 3sin2θ + 2cosθ)_π/6^5π/6 = (10π/3 - 3sin5π/3 + 2cos5π/6) -

-(2π/3 - 3sinπ/3 + 2cosπ/6) = 8π/3 + 3√3/2 - √3 + 3√3/2 - √3 = 8π/3 + √3 = 10.109

Answer is A