208k views
1 vote
Find the area of the region that is outside the curve r = 2 and inside the curve r = 4 cos θ. Your work must include the integral, but you may use your calculator to find the area to 3 decimal places.

1 Answer

1 vote

Answer:

4π/3 + 2√3 ≈ 7.653

Explanation:

Area between two polar curves is:

A = ∫ₐᵇ ½ (R² − r²) dθ

First, graph the curves. r = 2 is a circle with radius 2 and center (0,0). r = 4 cos θ is a circle with radius 2 and center (2,0).

Find where the curves intersect:

2 = 4 cos θ

1/2 = cos θ

θ = ±π/3

Between θ = -π/3 and θ = π/3, 4 cos θ > 2. So R = 4 cos θ and r = 2.

So the integral is:

A = ∫₋ᵖⁱ'³ ½ ((4 cos θ)² − 2²) dθ

A = ∫₋ᵖⁱ'³ ½ (16 cos²θ − 4) dθ

A = ∫₋ᵖⁱ'³ (8 cos²θ − 2) dθ

Using symmetry, we can write this as:

A = 2 ∫₀ᵖⁱ'³ (8 cos²θ − 2) dθ

A = ∫₀ᵖⁱ'³ (16 cos²θ − 4) dθ

Use power reduction formula:

A = ∫₀ᵖⁱ'³ (8 + 8 cos(2θ) − 4) dθ

A = ∫₀ᵖⁱ'³ (4 + 8 cos(2θ)) dθ

Integrate:

A = (4θ + 4 sin(2θ)) |₀ᵖⁱ'³

A = (4π/3 + 4 sin(2π/3)) − (0 + 0)

A = 4π/3 + 2√3

A ≈ 7.653

User Emmie
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories