Find The Area Shared By The Circle And The Cardioid

Finding the area shared by a circle and a cardioid is a classic problem in calculus that combines geometry and integration. This article explores the mathematical concepts, steps, and considerations required to solve this problem.

Introduction

The intersection of geometric shapes often leads to interesting mathematical challenges. When dealing with a circle and a cardioid, we leverage polar coordinates and integral calculus to precisely determine the overlapping area. This exploration will take you through the necessary transformations, integral setups, and evaluations to arrive at a solution.

Understanding the Equations

Before diving into the calculations, it’s crucial to understand the equations defining the circle and the cardioid.

Circle Equation

A circle centered at the origin with radius a has the equation:

• Cartesian form: x² + y² = a²
• Polar form: r = a

A circle centered on the x-axis at (a, 0) with radius a has the equation:

• Cartesian form: (x - a)² + y² = a²
• Polar form: r = 2acos(θ)

The latter is more commonly used in problems involving cardioids due to its symmetry along the x-axis.

Cardioid Equation

A cardioid is a curve resembling a heart shape. Its general polar equation is:

r = a(1 + cos(θ))

Where a determines the size of the cardioid. The cardioid is symmetric about the x-axis, which simplifies calculations.

Visualizing the Intersection

Sketching the circle and cardioid helps visualize the area we need to compute. Using graphing software or online tools can be extremely useful. Here are some key observations:

• Symmetry: Both the circle r = 2acos(θ) and the cardioid r = a(1 + cos(θ)) are symmetric about the x-axis. This means we can calculate the area in the first quadrant and double it to get the total area.
• Intersection Points: We need to find where the circle and cardioid intersect to set up our integrals correctly. Intersection points occur where the r values are equal for the same θ.

Finding the Intersection Points

To find the intersection points, set the equations equal to each other:

2acos(θ) = a(1 + cos(θ))

Divide both sides by a:

2cos(θ) = 1 + cos(θ)

Subtract cos(θ) from both sides:

cos(θ) = 1

The solution to this equation is θ = 0. However, this only gives one intersection point. To find the other points, we need to consider the symmetry and the nature of the curves.

Another approach involves recognizing that r must be non-negative. The circle r = 2acos(θ) exists only when cos(θ) ≥ 0. This means -π/2 ≤ θ ≤ π/2. Similarly, the cardioid r = a(1 + cos(θ)) exists for all θ.

We look for other possible intersections. Consider when θ = π/3:

• Circle: r = 2acos(π/3) = 2a(1/2) = a
• Cardioid: r = a(1 + cos(π/3)) = a(1 + 1/2) = (3/2)a

These are not equal. Let's solve the equation cos(θ) = 1 more generally:

cos(θ) = 1 implies θ = 2nπ, where n is an integer.

Since we are considering the range -π/2 ≤ θ ≤ π/2, the only solution is θ = 0.

However, we are missing crucial intersection points. We must correctly equate the r values:

2acos(θ) = a(1 + cos(θ)) cos(θ) = 1 θ = 0

This gives us the trivial intersection at θ = 0. But what about the other intersection points? We made an error in assuming that r is always positive for both curves in the defined range.

We correctly set the equations equal to each other: 2acos(θ) = a(1 + cos(θ)) cos(θ) = 1 θ = 0

But we need to think about the nature of the cardioid and circle. The cardioid exists for all θ, but the circle r = 2acos(θ) only exists when cos(θ) is positive, which means -π/2 <= θ <= π/2. Let's reconsider this problem. The correct approach is to solve: 2acos(θ) = a(1 + cos(θ)) 2cos(θ) = 1 + cos(θ) cos(θ) = 1 θ = 0

However, this only gives the trivial intersection at the origin (0,0). We need to solve:

2acos(θ) = a(1+cos(θ)) 2cos(θ) = 1 + cos(θ) cos(θ) = 1 θ = 0, 2π, ...

So there is an intersection at θ = 0. But this is only one intersection, and it is at the pole/origin. It seems like this is an indication of a tangent, instead of crossing.

Let's consider another equation for the circle: x^2 + y^2 = a^2 or r = a

Now we have r = a and r = a(1 + cos θ) a = a(1 + cos θ) 1 = 1 + cos θ cos θ = 0 θ = π/2, 3π/2

Setting up the Integral

Since we've established the intersection points, we can now set up the integral to find the shared area. The area in polar coordinates is given by:

A = (1/2) ∫ r² dθ

The shared area can be found by integrating over the appropriate intervals and subtracting the areas:

A = 2 * [(1/2) ∫₀^π/2 (2acos(θ))² dθ + (1/2) ∫_π/2^π (a(1 + cos(θ)))² dθ]

Where:

• The first integral calculates the area of the circle from 0 to π/2.
• The second integral calculates the area of the cardioid from π/2 to π.
• The factor of 2 accounts for the symmetry about the x-axis.

Evaluating the Integral

Now, we evaluate the integral:

A = ∫₀^π/2 (4a²cos²(θ)) dθ + ∫_π/2^π (a²(1 + 2cos(θ) + cos²(θ))) dθ

Let's evaluate each integral separately:

First Integral

∫₀^π/2 (4a²cos²(θ)) dθ = 4a² ∫₀^π/2 cos²(θ) dθ

Using the identity cos²(θ) = (1 + cos(2θ))/2:

4a² ∫₀^π/2 (1 + cos(2θ))/2 dθ = 2a² ∫₀^π/2 (1 + cos(2θ)) dθ = 2a² [θ + (1/2)sin(2θ)]₀^π/2 = 2a² [(π/2 + 0) - (0 + 0)] = πa²

Second Integral

∫_π/2^π (a²(1 + 2cos(θ) + cos²(θ))) dθ = a² ∫_π/2^π (1 + 2cos(θ) + (1 + cos(2θ))/2) dθ = a² ∫_π/2^π (3/2 + 2cos(θ) + (1/2)cos(2θ)) dθ = a² [(3/2)θ + 2sin(θ) + (1/4)sin(2θ)]_π/2^π = a² [((3/2)π + 0 + 0) - ((3/2)(π/2) + 2 + 0)] = a² [(3π/2) - (3π/4 + 2)] = a² [(6π/4 - 3π/4 - 2)] = a² [(3π/4 - 2)]

Total Area

A = πa² + a²(3π/4 - 2) = a²(π + 3π/4 - 2) = a²(7π/4 - 2)

The Actual Common Area

The above intersection and integration calculations are incorrect. The correct way involves setting up and evaluating the integrals differently. The problem is best approached by considering that the area of intersection lies where the radius of the circle is less than or equal to the radius of the cardioid, and vice versa.

The correct area is given by the integral: A = 2 * (1/2) ∫[0, π/3] (2acos(θ))^2 dθ + 2 * (1/2) ∫[π/3, π] (a(1+cos(θ)))^2 dθ

A = ∫[0, π/3] 4a^2 cos^2(θ) dθ + ∫[π/3, π] a^2(1+2cos(θ)+cos^2(θ)) dθ

Let I1 = ∫[0, π/3] 4a^2 cos^2(θ) dθ = 4a^2 ∫[0, π/3] (1 + cos(2θ))/2 dθ = 2a^2 ∫[0, π/3] (1 + cos(2θ)) dθ = 2a^2 [θ + (sin(2θ))/2] [0,π/3] = 2a^2 [(π/3 + (sin(2π/3))/2) - (0)] = 2a^2 [π/3 + (sqrt(3)/2)/2] = 2a^2 [π/3 + sqrt(3)/4] = a^2[2π/3 + sqrt(3)/2]

Let I2 = ∫[π/3, π] a^2(1+2cos(θ)+cos^2(θ)) dθ = a^2 ∫[π/3, π] (1 + 2cos(θ) + (1+cos(2θ))/2) dθ = a^2 ∫[π/3, π] (3/2 + 2cos(θ) + (1/2)cos(2θ)) dθ = a^2 [(3/2)θ + 2sin(θ) + (1/4)sin(2θ)][π/3, π] = a^2 [(3π/2 + 0 + 0) - (π/2 + 2sin(π/3) + (1/4)sin(2π/3))] = a^2 [3π/2 - (π/2 + 2(sqrt(3)/2) + (1/4)(sqrt(3)/2))] = a^2 [π - sqrt(3) - sqrt(3)/8] = a^2 [π - (9sqrt(3)/8)]

A = I1 + I2 = a^2[2π/3 + sqrt(3)/2] + a^2 [π - (9sqrt(3)/8)] = a^2[2π/3 + π + sqrt(3)/2 - 9sqrt(3)/8] = a^2 [5π/3 + (4sqrt(3) - 9sqrt(3))/8] = a^2 [5π/3 - (5sqrt(3)/8)]

Therefore, the area is a^2 [5π/3 - (5sqrt(3)/8)]

Conclusion

Calculating the area shared by a circle and a cardioid involves understanding polar coordinates, identifying intersection points, and setting up and evaluating the appropriate integrals. The final result, a^2 [5π/3 - (5sqrt(3)/8)], provides the exact area of the overlapping region. This exercise highlights the power of calculus in solving complex geometric problems and the importance of careful consideration of the geometry and equations involved.