Skip to main content

Double integral r sinθ over cardioid

Short answer

Integrate r sinθ dr dθ with r from 0 to a(1+cosθ), θ from 0 to π. The r-integral yields a²(1+cosθ)²/2; substitute t=1+cosθ to get 4a²/3.

Evaluate: ∬ r sin θ dr dθ over the area of the cardioid r = a(1+cos θ) above the initial line.

Region. The part of the cardioid r = a(1 + cos θ) above the initial line (θ = 0) is traced for θ from 0 to π. For each θ, r runs from 0 to a(1 + cos θ).

Reading. The integrand is taken exactly as printed, r sin θ, with the integration element dr dθ.

I = ∫₀^π ∫₀^{a(1+cosθ)} r sinθ dr dθ

r-integration

∫₀^{a(1+cosθ)} r dr = a²(1 + cos θ)²/2

θ-integration

Put t = 1 + cos θ, dt = −sin θ dθ. When θ = 0, t = 2; when θ = π, t = 0.

I = (a²/2) ∫₀^π (1 + cos θ)² sin θ dθ
  = (a²/2) ∫₀² t² dt
  = (a²/2)(8/3)
  = 4a²/3

Answer: 4a²/3.

If the area element r dr dθ were also included (that is, the integral of y = r sin θ over the area), the value would be 4a³/3. With the integral as printed, the answer is 4a²/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 2 questions →