For the polar curves r = 2 and r = sin feda?

Question

There are 3 parts to the question please help with as much as you can thanks!
a) Find the area that is inside r = 4sin feda and outside r = 2
b) Find the length of r = 4sin feda and when pie/4 < = Pie < = 3 pie /4 ( < = is smaller than and equal to ).
c) Find the line tangent to r = 4sin Feda at the point where Feda = Pie/6

Answer 1

a) r = 4sinθ and r = 2 intersect when
4sinθ = 2
θ = π/6, 5π/6
dA = rdrdθ
Integrating r between 2 and 4sinθ,
dA = [8sin^2θ - 2]dθ
dA = [4 - 4sin(2θ) - 2]dθ
dA = [2 - 4sin(2θ)]dθ
A = [2θ + 2cos(2θ)] |π/6 to 5π/6
A = 2{[5π/6 + cos(5π/3)] - [π/6 + cos(π/3)]}
A = 2[5π/6 + 1/2- π/6 + 1/2]
A = 2(π + 1) ≈ 8.283185

b) ds = √(dr^2 + r^2dθ^2)
ds = [√((dr/dθ)^2 + r^2)]dθ
dr/dθ = 4cosθ
ds = [√((4cosθ)^2 + (4sinθ)^2)]dθ
ds = 4dθ
s = 4(3π/4 - π/4) ≈ 6.283185

c) y = rsinθ = 4sin^2θ, y(π/6) = 1
x = rcosθ = 4sinθcosθ, x(π/6) = √3
dy = [4sinθcosθ]dθ
dx = [4cos^2θ - 4sin^2θ]dθ
dy/dx = (√3)/(3 - 1) = (1/2)√3
y - 1 = ((1/2)√3) (x - √3)
y = (x/2)√3 - 1/2