Volume or surface area of cycloid about base

Cycloid: x = a(θ + sin θ), y = a(1 + cos θ). One arch runs from θ = −π to π, with cusps at (∓πa, 0) on the x-axis, so the base is the x-axis. We have dx/dθ = a(1 + cos θ) = 2a cos²(θ/2) and dy/dθ = −a sin θ. The arc element is

ds = √(x′² + y′²) dθ = 2a cos(θ/2) dθ.

Also y = 2a cos²(θ/2).

Volume about the base (x-axis)

V = π ∫ y² dx = π ∫_(−π)^π a²(1+cosθ)² · a(1+cosθ) dθ
  = πa³ ∫_(−π)^π (1 + cosθ)³ dθ
  = 8πa³ ∫_(−π)^π cos⁶(θ/2) dθ

Put θ/2 = u: ∫(−π)^π cos⁶(θ/2) dθ = 2∫(−π/2)^(π/2) cos⁶u du = 4∫₀^(π/2) cos⁶u du = 4 · 5π/32 = 5π/8.

V = 8πa³ · 5π/8 = 5π²a³ cubic units.

Surface area about the base

S = 2π ∫ y ds = 2π ∫_(−π)^π 2a cos²(θ/2) · 2a cos(θ/2) dθ
  = 8πa² ∫_(−π)^π cos³(θ/2) dθ

With θ/2 = u: ∫ cos³(θ/2) dθ = 2∫_(−π/2)^(π/2) cos³u du = 4∫₀^(π/2) cos³u du = 4 · 2/3 = 8/3.

S = 8πa² · 8/3 = 64πa²/3 square units.