Volume of cycloid about tangent at vertex

Reading of the question. The curve x = a(θ + sin θ), y = a(1 − cos θ) has its cusps at y = 2a (at θ = ±π) and its vertex at the origin, where the tangent is the x-axis (y = 0). The usual textbook problem (answer 7π²a³) asks for the solid formed when the area between one arch and the line joining its cusps is revolved about the tangent at the vertex. We solve that version.

Setup. One arch runs for θ from −π to π, and dx = a(1 + cos θ) dθ. The area enclosed between the arch and the cusp line y = 2a is revolved about y = 0. The strip at x has outer radius 2a and inner radius y, so

V = π ∫ (4a² − y²) dx
  = π ∫_(−π)^π [4a² − a²(1 − cosθ)²] · a(1 + cosθ) dθ

Split the integral:

• ∫ 4a² · a(1 + cos θ) dθ over (−π, π) = 4a³ · 2π = 8πa³.
• ∫ a²(1 − cos θ)² a(1 + cos θ) dθ = a³ ∫ (1 − cos θ) sin²θ dθ = a³ [∫ sin²θ dθ − ∫ cos θ sin²θ dθ] = a³ (π − 0) = πa³.

V = π (8πa³ − πa³)

V = 7π²a³ cubic units.

(If instead the region between the arch and the tangent y = 0, bounded by x = ±πa, is revolved about y = 0, the volume is π∫y² dx = π²a³. The standard result 7π²a³ is for the region between the arch and the cusp line.)