Calculus Question

Question

Find the area of the region inside the cardiod r=2+2cos∅ and outside the circle r=3

Answer 1

The general area integral is dydx but in this case is is rdrda ( a = theta)

the limits for this integral are 3<r<2+2cos(a) and -pi/3<a<pi/3

The integral result is the area which is 4.653

Answer 2

Hello Divya,

First, we will find the points of intersection of the curves given by the 2 polar equations. Setting the r values equal, we have

2 + 2cosθ = 3

2cosθ = 1

cosθ = 1/2

There are two solutions, one in the first quadrant and one in the 4th quadrant

θ = π/3 and θ = 5π/3.

The angle 5π/3 is equivalent (coterminal) to the angle -π/3, and it will be more convenient to use -π/3 in setting up the integral. Suppose the functions r₂ = f(θ) and r₁ = g(θ) define two curves (in polar coordinates). Assume that curve given by r₂ is outside the curve given by r₁ for the interval of θ-values (θ₁,θ₂). That is, suppose that r₂ > r₁ if θ₁ < θ < θ₂. Then

A = ∫(θ₁ to θ₂) (1/2)[r₂² - r₁²] dθ

gives the area between the two curves as θ goes from θ₁ to θ₂. [That is, the area outside of r₁ = g(θ) and inside r₂ = f(θ).] In our case,

A = ∫(-π/3 to π/3) (1/2)[(2 + 2cosθ)² - 3²] dθ

A = (1/2)∫(-π/3 to π/3) [(2 + 2cosθ)² - 3²] dθ

A = (1/2)∫(-π/3 to π/3) [4cos²(θ) + 8cosθ - 5] dθ

Using the identity cos²θ = (1 + cos(2θ))/2 and doing some algebra, this becomes

A = (1/2)∫(-π/3 to π/3) [2cos(2θ) + 8cosθ - 3] dθ.

Completing the integral, we have

A = (1/2){[sin(2θ) + 8sinθ - 3θ] [from -π/3 to π/3]}

A = (1/2)[9√3 – 2π] = (9/2)√3 - π

Hope that helps! Let me know if you need any further explanation.

William