Compute the area A = (1/2) ∫_{-π/2}^{π/2} (2 cos θ)^2 dθ for the area enclosed by r = 2 cos θ over that interval.

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Multiple Choice

Compute the area A = (1/2) ∫_{-π/2}^{π/2} (2 cos θ)^2 dθ for the area enclosed by r = 2 cos θ over that interval.

Explanation:
The area in polar coordinates is given by A = (1/2) ∫ r^2 dθ. Here r = 2 cos θ, so A = (1/2) ∫_{-π/2}^{π/2} (2 cos θ)^2 dθ = 2 ∫_{-π/2}^{π/2} cos^2 θ dθ. Use the identity cos^2 θ = (1 + cos 2θ)/2 to get A = ∫_{-π/2}^{π/2} (1 + cos 2θ) dθ = [θ + (1/2) sin 2θ]_{-π/2}^{π/2}. The sine terms vanish at the endpoints, and the θ terms contribute π/2 − (−π/2) = π. So A = π. Geometrically, r = 2 cos θ describes the circle x^2 + y^2 = 2x, which is a circle of radius 1 centered at (1, 0). The interval [-π/2, π/2] traces that entire circle once, so its area is π.

The area in polar coordinates is given by A = (1/2) ∫ r^2 dθ. Here r = 2 cos θ, so A = (1/2) ∫{-π/2}^{π/2} (2 cos θ)^2 dθ = 2 ∫{-π/2}^{π/2} cos^2 θ dθ.

Use the identity cos^2 θ = (1 + cos 2θ)/2 to get A = ∫{-π/2}^{π/2} (1 + cos 2θ) dθ = [θ + (1/2) sin 2θ]{-π/2}^{π/2}. The sine terms vanish at the endpoints, and the θ terms contribute π/2 − (−π/2) = π. So A = π.

Geometrically, r = 2 cos θ describes the circle x^2 + y^2 = 2x, which is a circle of radius 1 centered at (1, 0). The interval [-π/2, π/2] traces that entire circle once, so its area is π.

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