The area of the region outside r=5+5sinθ, but inside r=15sinθ is 78.55 unit².
How the area of the region is calculated.
Given
Outside radius = 5 + 5 sinθ θ
Inside radius = 15sinθ
Point where the radii intersect is
r₁ = r₂
5 + 5sinθ = 15sinθ
5 = 15sinθ - 5sinθ
5 = 10sinθ
sinθ = 5/10
sinθ = 1/2
θ = sin⁻¹(0.5)
θ = 30⁰, 150⁰ (Sine is positive in first and second quadrant)
r₂ = 15sinθ and r₁ = 5 + 5sinθ
Area of region = 1/2∫[a,b](r₂²- r₁²)dθ
where
a = 30⁰ and b = 150⁰
A = 1/2∫₃₀¹⁵⁰[(15sinθ)² - (5 + 5sinθ)²]dθ
= 1/2∫₃₀¹⁵⁰[225sin²θ - (25 + 50sinθ + 25sin²θ)]dθ
=1/2∫₃₀¹⁵⁰[200sin²θ - 50sinθ -25)]dθ
Sin²θ = 1 - cos2θ/2
A = 1/2∫₃₀¹⁵⁰[100(1 - cos2θ) - 50sinθ -25)]dθ
= 1/2∫₃₀¹⁵⁰[100 - 100cos2θ - 50sinθ -25)]dθ
Integrate
= 1/2[100θ - 50sinθ - 25θ + 50cosθ]₃₀¹⁵⁰
= 1/2[75θ - 50sinθ + 50cosθ]₃₀¹⁵⁰
Substitute the limits
= 1/2{[75(150) - 50sin(150) + 50cos(150)] - [75(30) - 50sin(30) + 50cos(30)]}
= 1/2{[75(0.5) - 50(0.5) + 50(-0.866)] - [75(0.5) - 50(0.5) + 50(-0.866)]}
=78.55 unit²
The area of the region outside r=5+5sinθ, but inside r=15sinθ is 78.55 unit².