Calculate Area of Circle by Integration

What is Integration?

Integration is a fundamental operation in calculus that finds the area under a curve, the accumulation of quantities, and the reversal of differentiation. It's the inverse process of differentiation. In practical terms, integration allows us to calculate areas, volumes, and other quantities that can be expressed as the limit of a sum.

The definite integral of a function f(x) from a to b, denoted as ∫[a,b] f(x) dx, represents the signed area between the curve y = f(x) and the x-axis from x = a to x = b. The result is a number that gives the net area between the curve and the x-axis.

Area of a Circle

The area of a circle is a well-known geometric property that can be calculated using the simple formula A = πr², where r is the radius of the circle. However, this formula is typically derived using geometric methods. Calculating the area using integration provides an alternative approach that connects calculus with geometry.

To find the area of a circle using integration, we can consider the circle as the area under the curve of a semicircle. By integrating the function that describes the upper half of the circle, we can find the total area.

Calculating Area Using Integration

The equation of a circle centered at the origin with radius r is x² + y² = r². Solving for y gives us y = ±√(r² - x²). To find the area of the upper semicircle, we can integrate the positive square root function from -r to r.

This integral can be evaluated using trigonometric substitution. Let x = r sinθ, then dx = r cosθ dθ. The limits change as follows:

• When x = -r, θ = -π/2
• When x = r, θ = π/2

Substituting these into the integral gives:

Using the identity cos²θ = (1 + cos2θ)/2, we can simplify the integral:

Evaluating this at the limits gives:

Since the area of a full circle is twice the area of a semicircle, we get the familiar formula A = πr².

Worked Example

Let's calculate the area of a circle with radius r = 5 units using integration.

1. Write the integral for the upper semicircle: ∫[-5,5] √(25 - x²) dx
2. Use trigonometric substitution: x = 5 sinθ, dx = 5 cosθ dθ
3. Change the limits: θ goes from -π/2 to π/2
4. Substitute into the integral: ∫[-π/2,π/2] √(25 - 25 sin²θ) * 5 cosθ dθ = ∫[-π/2,π/2] 5 cosθ * 5 cosθ dθ = 25 ∫[-π/2,π/2] cos²θ dθ
5. Use the identity cos²θ = (1 + cos2θ)/2: 25 ∫[-π/2,π/2] (1 + cos2θ)/2 dθ = (25/2) [θ + (sin2θ)/2] from -π/2 to π/2
6. Evaluate at the limits: (25/2) [(π/2 + 0) - (-π/2 + 0)] = (25/2) [π] = (25π)/2 = 12.5π
7. Multiply by 2 for the full circle: 2 * 12.5π = 25π

The calculated area is 25π square units, which matches the standard formula A = πr² for r = 5.