What is the area enclosed by the polar curve r = 1 + cos θ for θ from 0 to 2π?

What is the area enclosed by the polar curve r = 1 + cos θ for θ from 0 to 2π?

• A. π
• B. 2π
• C. (3/2)π
• D. (4/3)π

Answer: (C)

Area in polar coordinates is found by integrating one-half times r squared with respect to θ over the given interval. For r = 1 + cos θ, first square the radius: r^2 = (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ = 3/2 + 2 cos θ + (1/2) cos 2θ. Then the enclosed area is A = (1/2) ∫ from 0 to 2π of [3/2 + 2 cos θ + (1/2) cos 2θ] dθ. The constant part gives (1/2) · (3/2) · (2π) = 3π/2, while the integrals of cos θ and cos 2θ over a full period vanish. So A = 3π/2. So the area enclosed is (3/2)π.

Area in polar coordinates is found by integrating one-half times r squared with respect to θ over the given interval. For r = 1 + cos θ, first square the radius: r^2 = (1 + cos θ)^2 = 1 + 2 cos θ + cos^2 θ = 3/2 + 2 cos θ + (1/2) cos 2θ.

Then the enclosed area is A = (1/2) ∫ from 0 to 2π of [3/2 + 2 cos θ + (1/2) cos 2θ] dθ. The constant part gives (1/2) · (3/2) · (2π) = 3π/2, while the integrals of cos θ and cos 2θ over a full period vanish. So A = 3π/2.

So the area enclosed is (3/2)π.