Use a double integral in polar coordinates to find the area of the region bounded on the inside by the circle of radius 1 and on the outside by the cardioid r=1(1+cos(θ))

Answer 1

To find the area of the region bounded on the inside by the circle of radius 1 and on the outside by the cardioid r = 1(1 + cos(θ)), we can set up a double integral in polar coordinates.

First, let's sketch the region we want to find the area of:

The circle of radius 1 is centered at the origin and the cardioid is also symmetric about the origin. The cardioid has a point at (0,1), which is the highest point on the curve, and it reaches the origin when θ = π.

To set up the double integral, we need to find the limits of integration for r and θ.

Let's find the value of θ when the cardioid intersects the circle of radius 1. Since r = 1(1 + cos(θ)) for the cardioid, we can substitute r = 1 into this equation:

1 = 1(1 + cos(θ))
1 = 1 + cos(θ)
cos(θ) = 0
θ = π/2 or 3π/2

So the cardioid intersects the circle of radius 1 at θ = π/2 and θ = 3π/2.

Also, since we want to find the area inside the circle, the cardioid will always have a larger value of r compared to the circle. Therefore, the limits of integration for r will be from 0 to 1(1 + cos(θ)).

Now we can set up the double integral:

A = ∫∫ r dr dθ

The limits of integration for r are from 0 to 1(1 + cos(θ)), and for θ, they are from π/2 to 3π/2.

A = ∫[π/2 to 3π/2] ∫[0 to 1(1 + cos(θ))] r dr dθ

Now, we can integrate with respect to r first:

A = ∫[π/2 to 3π/2] [(1/2)r^2] (from 0 to 1(1 + cos(θ))) dθ
= ∫[π/2 to 3π/2] [(1/2)(1 + cos(θ))^2] dθ

Next, we integrate with respect to θ:

A = 1/2 ∫[π/2 to 3π/2] (1 + 2cos(θ) + cos^2(θ)) dθ
= 1/2 [θ + 2sin(θ) + (θ/2 + (sin(2θ)/4) + (1/4)(θ + sin(θ)cos(θ)))| from π/2 to 3π/2

Plugging in the limits of integration:

A = 1/2 [3π/2 + 2sin(3π/2) + (3π/4 + (sin(3π)/4) + (1/4)(3π/2 + sin(3π/2)cos(3π/2)))] - 1/2 [π/2 + 2sin(π/2) + (π/4 + (sin(π)/4) + (1/4)(π/2 + sin(π/2)cos(π/2)))]

Simplifying further, we get:

A = π/4 + sin(3π/2)/4 + sin(π)/8 - sin(π/2)/4

Since sin(3π/2) = sin(π) = -1 and sin(π/2) = 1, the expression becomes:

A = π/4 - 1/4 + 1/8 + 1/4 = π/2 - 1/8

Therefore, the area of the region bounded on the inside by the circle of radius 1 and on the outside by the cardioid r = 1(1 + cos(θ)) is π/2 - 1/8.