area of polar curves?

Question

find the area of the region which is inside the polar curve r=6cos(theta) and the outside polar curve is r=4-2cos(theta).
what is the area?

Answer 1

The two curves intercept at t =+-pi/3
as 4-2cos t =6 cos t==> cos t=1/2
You have to calculate the integral S =1/2Int( 8cos t-4)^2 dt
taken between -pi3 and pi/3
=8Int( 4cos^2t-4cost+1) dt = 8(sin2t+2t-4sint+t) (-pi/3,pi/3)
you can do the substitutions

Answer 2

hit upon the aspect to the area it is interior the polar curve r = 6cos? and the exterior polar curve r = 4 - 2cos?. First hit upon the criteria of intersection. r = 6cos? = 4 - 2cos? 8cos? = 4 cos? = a million/2 ? = ±?/3 A = area combine from -?/3 to ?/3. A = (a million/2)?[R² - r²]d? = (a million/2)?[(6cos?)² - (4 - 2cos?)²] d? = (a million/2)?[36cos²? - (16 - 16cos? + 4cos²?)] d? = (a million/2)?[32cos²? + 16cos? - 16] d? = ?[16cos²? + 8cos? - 8] d? = ?{8[a million + (cos 2?)] + 8cos? - 8} d? = ?{8(cos 2?) + 8cos?} d? = 4(sin 2?) + 8sin? | [Eval from -?/3 to ?/3] = [4(?3/2) + 8(?3/2)] - [4(-?3/2) + 8(-?3/2)] = 6?3 + 6?3 = 12?3

Answer 3

Find the area of the region which is inside the polar curve
r = 6cosθ and the outside polar curve r = 4 - 2cosθ.

First find the points of intersection.

r = 6cosθ = 4 - 2cosθ
8cosθ = 4
cosθ = 1/2
θ = ±π/3

A = area

Integrate from -π/3 to π/3.

A = (1/2)∫[R² - r²]dθ

= (1/2)∫[(6cosθ)² - (4 - 2cosθ)²] dθ

= (1/2)∫[36cos²θ - (16 - 16cosθ + 4cos²θ)] dθ

= (1/2)∫[32cos²θ + 16cosθ - 16] dθ

= ∫[16cos²θ + 8cosθ - 8] dθ

= ∫{8[1 + (cos 2θ)] + 8cosθ - 8} dθ

= ∫{8(cos 2θ) + 8cosθ} dθ

= 4(sin 2θ) + 8sinθ | [Eval from -π/3 to π/3]

= [4(√3/2) + 8(√3/2)] - [4(-√3/2) + 8(-√3/2)]

= 6√3 + 6√3 = 12√3