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))
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.
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 use a double integral in polar coordinates.
First, let's graph the region to get a visual understanding of what we are working with.
The equation of the circle is r = 1, and the equation of the cardioid is r = 1(1 + cos(θ)).
We can see that the cardioid intersects the circle at two points: θ = 0 and θ = π. To find the limits of integration, we need to determine the values of θ that define the region.
Therefore, the limits of integration for θ will be [0, π].
The limits of integration for r will be [0, 1(1 + cos(θ))].
Now, we can set up the double integral to find the area of the region:
A = ∬ R dA,
where R is the region defined by the given equations.
The area element in polar coordinates is given by dA = r dr dθ.
Substituting this expression for dA, we have:
A = ∫[0,π] ∫[0,1(1+cos(θ))] r dr dθ.
Now, we can evaluate this double integral:
A = ∫[0,π] ∫[0,1(1+cos(θ))] r dr dθ.
Simplifying the integral, we have:
A = ∫[0,π] (1/2) r^2 |[0,1(1+cos(θ))]| dθ.
Evaluating the integral and substituting the limits, we have:
A = (1/2) ∫[0,π] (1/2) (1(1+cos(θ)))^2 dθ.
Expanding the expression, we have:
A = (1/2) ∫[0,π] (1/2) (1 + 2cos(θ) + cos^2(θ)) dθ.
Using the trigonometric identity cos^2(θ) = (1/2)(1 + cos(2θ)), we can simplify further:
A = (1/2) ∫[0,π] (1/2) (1 + 2cos(θ) + (1/2)(1 + cos(2θ))) dθ.
Expanding and simplifying again, we have:
A = (1/2) ∫[0,π] (3/2) + cos(θ) + (1/4)cos(2θ) dθ.
Integrating each term individually, we get:
A = (1/2) [(3/2)θ + sin(θ) + (1/8)sin(2θ)] |[0,π].
Substituting the limits, we have:
A = (1/2) [(3/2)π + sin(π) + (1/8)sin(2π)] - (1/2) [(3/2)0 + sin(0) + (1/8)sin(2(0))].
Simplifying further, we have:
A = (1/2) [(3/2)π + 0 + 0] - (1/2) [0 + 0 + 0].
Finally, we get:
A = (1/2) (3/2)π.
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 (3/4)π square units.