Find the area of the region which is inside the polar curve?

Question

r=4cos(@) and outside the curve r=3-2cos(@)....where @ equals theta or angle

Answer 1

Greetings,

First find where the curves intersect, this will give the limits of integration.

4cosθ = 3 - 2cosθ
6cosθ = 3
cosθ = 1/2
θ = π/3 or 5π/3

A = 1/2∫(r of outer curve)^2 - (r of inner curve)^2 ) dθ
= 1/2∫[16(cosθ)^2 - (3 - 2cosθ)^2] dθ
= 1/2∫[12(cosθ)^2 +12cosθ - 9] dθ

since cosine is symmetric around θ, the area is twice the integral with limits of integration from 0 to π/3

= [12(θ/2 +sinθcosθ/2) +12sinθ - 9θ] now evaluate between the limits

= 15/2sqrt(3) - π

Regards

Answer 2

To determine the area of the specified region, you first have to determine its bounds.

Where are the points where the two curves intersect? i.e. where 4 cos(@) = 3 - 2 cos(@)

To compute the area in the region, we are going to have integrate A(@)d@. Which of the intersection points are the start of the integration and which are the end of the integration

(That is, for some intervals, 4 cos(@) < 3 - 2 cos(@) so the area between the curves is outside 4 cos(@) rather than inside. W don't want to include those intervals in our integration)

Then we need to determine A(@). The outer boundary is
Ro(@) = 4 cos (@)

and the inner boundary is:
Ri(@) = 3 - 2 cos (@)

So A(@) = (1/2)(Ro(@) - Ri(@))