Compute the area enclosed by one loop of r = 1 + cos θ.

Compute the area enclosed by one loop of r = 1 + cos θ.

• A. π
• B. 2π
• C. 3π/2
• D. π/2

Answer: (C)

In polar coordinates, the area swept by a curve r = f(θ) as θ runs over an interval is A = (1/2) ∫ r^2 dθ over that interval. For r = 1 + cos θ, the loop is completed as θ goes from 0 to 2π (r is nonnegative and the cardioid closes at θ = π). So compute the area as A = (1/2) ∫_0^{2π} (1 + cos θ)^2 dθ. Expand and integrate: (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ. Over 0 to 2π, ∫ 1 dθ = 2π, ∫ 2 cos θ dθ = 0, and ∫ cos^2 θ dθ = ∫ (1 + cos 2θ)/2 dθ = π. Sum gives 2π + π = 3π. Multiply by 1/2 to obtain A = (1/2)·3π = 3π/2. Therefore, the area enclosed by one loop is 3π/2.

In polar coordinates, the area swept by a curve r = f(θ) as θ runs over an interval is A = (1/2) ∫ r^2 dθ over that interval. For r = 1 + cos θ, the loop is completed as θ goes from 0 to 2π (r is nonnegative and the cardioid closes at θ = π). So compute the area as A = (1/2) ∫_0^{2π} (1 + cos θ)^2 dθ.

Expand and integrate: (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ. Over 0 to 2π, ∫ 1 dθ = 2π, ∫ 2 cos θ dθ = 0, and ∫ cos^2 θ dθ = ∫ (1 + cos 2θ)/2 dθ = π. Sum gives 2π + π = 3π. Multiply by 1/2 to obtain A = (1/2)·3π = 3π/2.

Therefore, the area enclosed by one loop is 3π/2.