Question

Find the area of the region that is inside r=3cos(theta) and outside r=2-cos(theta). Sketch the curves.​

Answer 1

Answer:

3√3

Step-by-step explanation:

r = 3 cos θ

r = 2 - cos θ

First, find the intersections.

3 cos θ = 2 - cos θ

4 cos θ = 2

cos θ = 1/2

θ = -π/3, π/3

We want the area inside the first curve and outside the second curve.  So R = 3 cos θ and r = 2 - cos θ, such that R > r.

Now that we have the limits, we can integrate.

A = ∫ ½ (R² - r²) dθ

A = ∫ ½ ((3 cos θ)² - (2 - cos θ)²) dθ

A = ∫ ½ (9 cos² θ - (4 - 4 cos θ + cos² θ)) dθ

A = ∫ ½ (9 cos² θ - 4 + 4 cos θ - cos² θ) dθ

A = ∫ ½ (8 cos² θ + 4 cos θ - 4) dθ

A = ∫ (4 cos² θ + 2 cos θ - 2) dθ

Using power reduction formula:

A = ∫ (2 + 2 cos(2θ) + 2 cos θ - 2) dθ

A = ∫ (2 cos(2θ) + 2 cos θ) dθ

Integrating:

A = (sin (2θ) + 2 sin θ) |-π/3 to π/3

A = (sin (2π/3) + 2 sin(π/3)) - (sin (-2π/3) + 2 sin(-π/3))

A = (½√3 + √3) - (-½√3 - √3)

A = 1.5√3 - (-1.5√3)

A = 3√3

The area inside of r = 3 cos θ and outside of r = 2 - cos θ is 3√3.

The graph of the curves is:

desmos.com/calculator/541zniwefe