Double integral r³ between two circles

Region. r = 2 sin θ and r = 4 sin θ are circles through the origin, with diameters 2 and 4 along the y-axis (centres (0,1) and (0,2)). The area between them is the region 2 sin θ ≤ r ≤ 4 sin θ, and θ runs from 0 to π (the circles start and end at the origin).

y
4 |    ___
  |  /  _  \
2 | |  / \  |   inner: r = 2 sinθ
  |  \ \_/ /    outer: r = 4 sinθ
  |____'-'____ x

Integral

I = ∫₀^π ∫_{2sinθ}^{4sinθ} r³ dr dθ
  = ∫₀^π [ r⁴/4 ]_{2sinθ}^{4sinθ} dθ
  = ∫₀^π (256 − 16)/4 · sin⁴θ dθ
  = 60 ∫₀^π sin⁴θ dθ

Using sin⁴θ = (3 − 4 cos 2θ + cos 4θ)/8, ∫₀^π sin⁴θ dθ = 3π/8.

I = 60 · 3π/8 = 45π/2

Answer: 45π/2.