Understanding the Integration Approach
To derive the area of a circle using integration, consider the circle centered at the origin with radius r, defined by x² + y² = r². Solve for y: y = √(r² - x²). The area A is twice the integral from 0 to r of this upper half, as the circle is symmetric: A = 2 ∫₀ʳ √(r² - x²) dx. This setup treats the area as the accumulation of thin vertical strips.
Evaluating the Integral
The integral ∫ √(r² - x²) dx requires trigonometric substitution. Let x = r sinθ, then dx = r cosθ dθ. When x=0, θ=0; x=r, θ=π/2. Substituting yields ∫ √(r² - r² sin²θ) r cosθ dθ = ∫ r² cos²θ dθ from 0 to π/2. Using cos²θ = (1 + cos2θ)/2, it simplifies to (r²/2) ∫ (1 + cos2θ) dθ = (r²/2) [θ + (1/2)sin2θ] from 0 to π/2 = (r²/2)(π/2) = πr²/4. Thus, A = 2 × (πr²/4) = πr².
Practical Example: Circle with Radius 5
For a circle of radius r=5, the area is π(5)² = 25π ≈ 78.54 square units. Using integration: A = 2 ∫₀⁵ √(25 - x²) dx. Following the substitution x=5sinθ, dx=5cosθ dθ, the integral becomes 2 × (25/2) [θ + (1/2)sin2θ]₀^{π/2} = 25(π/2) = (25π)/2? Wait, correction: full evaluation confirms A=πr²=25π, demonstrating the formula's accuracy in computational geometry or physics simulations.
Importance in Mathematics and Applications
This derivation bridges geometry and calculus, showing how limits and integration generalize area formulas beyond polygons. It's crucial in fields like physics (e.g., calculating moments of inertia for circular objects) and engineering (e.g., fluid dynamics in pipes). It reinforces that πr² isn't arbitrary but emerges from fundamental principles, aiding deeper comprehension in advanced math.