Démonstrations de Calcul Intégral

Aire du Cercle par Intégration

Apprenez comment les intégrales démontrent rigoureusement la formule A = πR² avec les coordonnées polaires, cartésiennes et les sommes de Riemann.

Calculus & Definite Integrals

Circle Area by Integration

Circle Radius (R)

units

Integration Method

Riemann Subdivisions (n): 16 strips

n = 4 (coarse) n = 16 n = 32 n = 64 (smooth)

Integral Formulation:

∫₀ᴿ 2πr dr = [ πr² ]₀ᴿ = πR²

The circle is accumulated as an infinite number of concentric thin rings of circumference 2πr and infinitesimal width dr.

Exact Integrated Area (πR²)

0 units²

Riemann Sum (n slices) 0

Convergence Accuracy 100%

Integral Strip Decomposition R = 0

Step-by-Step Proof:

Enter radius to evaluate the definite integral.

Why Use Calculus to Prove the Area of a Circle?

In elementary geometry, A = πr² is stated as a memorized formula. But where does it come from? How do mathematicians prove that the ratio of circumference to diameter (π) also governs the enclosed two-dimensional area?

Integral calculus provides the formal machinery to break a curved, continuous two-dimensional region into infinitely many infinitesimal elements (rings, wedges, or rectangular strips) and sum them rigorously.

Depending on your choice of coordinate system — concentric rings, polar double integrals, Cartesian trigonometric substitution, or Green's theorem contour integration — the math converges unequivocally on πR².

Comparison of Circle Area Integration Methods

Method	Differential Area (dA)	Integral Expression	Calculus Level
1. Concentric Shells / Rings	dA = 2πr dr	∫₀ᴿ 2πr dr = πR²	Single Variable (AP Calc AB)
2. Polar Coordinates	dA = r dr dθ	∫₀²ᵖᶦ ∫₀ᴿ r dr dθ = πR²	Multivariable (Calculus III)
3. Cartesian Trig Substitution	dA = 2√(R² - x²) dx	4 ∫₀ᴿ √(R² - x²) dx = πR²	Single Variable (AP Calc BC)
4. Green's Theorem Line Integral	½(x dy - y dx)	½ ∮_C (x dy - y dx) = πR²	Vector Calculus (Stokes' Law)

4 Step-by-Step Mathematical Proofs

Review each rigorous proof step-by-step from foundational axioms to final evaluation.

Proof 1: Concentric Rings (Shell Method)

Partition the circle into thin concentric circular ribbons of circumference 2πr and thickness dr. Integrate from r = 0 to r = R.

∫₀ᴿ 2πr dr = 2π [ ½ r² ]₀ᴿ = πR²

Proof 2: Polar Double Integral (Jacobian)

In polar coordinates x = r cos θ, y = r sin θ, the area element is dA = r dr dθ. Integrate radius from 0 to R and angle from 0 to 2π.

( ∫₀²ᵖᶦ dθ ) × ( ∫₀ᴿ r dr ) = 2π × ½ R² = πR²

Proof 3: Cartesian Trigonometric Substitution

From circle equation x² + y² = R², upper boundary is y = √(R² - x²). Using substitution x = R sin θ over 4 quadrants.

4 ∫₀^(π/2) R² cos²θ dθ = 4R² (π / 4) = πR²

Proof 4: Green's Theorem Line Integral

Green's Theorem evaluates area via boundary curve line integral: Area = ½ ∮ (x dy - y dx) along x = R cos t, y = R sin t.

½ ∫₀²ᵖᶦ R² dt = ½ R² (2π) = πR²

Rigorous Proof: Cartesian Trig Substitution Walkthrough

Step-by-step evaluation of 4 ∫₀ᴿ √(R² - x²) dx:

1. Cartesian Integral: Area = 4 ∫₀ᴿ √(R² - x²) dx

2. Let x = R sin θ, then dx = R cos θ dθ

3. Limit change: When x = 0 → θ = 0; when x = R → θ = π/2

4. Radical simplification: √(R² - R² sin²θ) = R cos θ

5. Transformed Integral: 4 ∫₀^(π/2) (R cos θ)(R cos θ dθ) = 4R² ∫₀^(π/2) cos²θ dθ

6. Half-Angle Identity: cos²θ = (1 + cos 2θ) / 2

7. Antiderivative: 4R² [ θ/2 + (sin 2θ)/4 ]₀^(π/2) = 4R² [ π/4 + 0 ]

8. Final Result: 4R² × (π / 4) = πR² [Q.E.D.]

Riemann Sum Numerical Convergence Table (R = 10)

How finite rectangular strips converge to the exact analytical calculus value πR² = 314.1593 as strip count n increases.

Number of Strips (n)	Left Riemann Sum	Midpoint Riemann Sum	Trapezoidal Rule	Percentage Error
4	341.42	315.65	291.42	+0.47%
8	328.52	314.54	303.52	+0.12%
16	321.49	314.26	308.99	+0.032%
64	316.02	314.16	312.89	+0.002%
1,000	314.28	314.1593	314.08	< 0.0001%
∞ (Analytical)	314.1593	314.1593	314.1593	0.0000% (Exact)

4 Solved Advanced Calculus Applications

Exam-style calculus problems demonstrating how circle integration extends to rotational physics and 3D solids.

Problem 1: Rotational Dynamics

Mass Moment of Inertia of a Uniform Circular Disc

Find the moment of inertia I_z of a uniform flat disc of mass M and radius R about its perpendicular central axis:

Mass per unit area: σ = M / (πR²)

Ring element mass: dm = σ(2πr dr)

I_z = ∫₀ᴿ r² dm = 2πσ ∫₀ᴿ r³ dr = 2πσ [ r⁴ / 4 ]₀ᴿ

Result: I_z = ½ M R²

Problem 2: Centroid Mechanics

Centroid of a Semicircular Region via Double Integral

Find the vertical center of mass ȳ of a semicircle of radius R:

Semicircle Area A = ½ πR²

First Moment Q_x = ∬ y dA = ∫₀^π ∫₀ᴿ (r sin θ)(r dr dθ)

Q_x = (2R³) / 3

ȳ = Q_x / A = [(2R³)/3] / [½ πR²] = (4R) / (3π) ≈ 0.4244R

Problem 3: Solid of Revolution

Volume of a Sphere via Circular Disk Slices

Derive the volume of a sphere of radius R by revolving y = √(R² - x²) around the x-axis:

Differential slice volume: dV = π y² dx = π(R² - x²) dx

Volume = ∫₋ᵣᴿ π(R² - x²) dx = 2π ∫₀ᴿ (R² - x²) dx

Antiderivative: 2π [ R²x - x³/3 ]₀ᴿ = 2π [ (2/3) R³ ]

Sphere Volume = (4/3) π R³

Problem 4: Fluid Work Integral

Work Required to Pump Out a Hemispherical Tank

Calculate work required to pump water of weight density w to top rim of radius R tank:

Circular water slice at depth y has radius r = √(R² - y²)

Slice Volume: dV = π(R² - y²) dy

Work Element: dW = w · y · dV = w π y(R² - y²) dy

Total Pumping Work = (w π R⁴) / 4

Common Pitfalls on Calculus Integration Exams

Forgetting the Jacobian 'r' in Polar Coordinates

Wrong: ∬ dr dθ. Correct: ∬ r dr dθ. Without the r factor, you integrate circumference 2πR, not area! The r factor converts dθ to arc length.

Forgetting to Multiply by 4 in Cartesian Integrals

Integrating ∫₀ᴿ √(R² - x²) dx yields only Quadrant 1 (πR² / 4). You must multiply by 4 to capture the entire circle.

Pro Tip: Derivative of Area is Circumference!

Notice that d/dr (πr²) = 2πr! The instantaneous rate of change of a circle's area as its radius expands is exactly its perimeter circumference.

Pro Tip: Triangle Unrolling Visualization

Cut the concentric shells from center to rim and lay them flat. They form a triangle of base 2πR and height R. Area = ½ × 2πR × R = πR².

Scientific & Engineering Applications of Circle Integration

Quantum Mechanics

Integrating radial wavefunctions |ψ|² 4πr² dr to find total probability density of an electron around a nucleus.

Electromagnetism

Applying Ampere's Law and Gauss's Law: ∮ B·dl = μ₀ I_enc over circular loop flux integrations.

Finite Element Analysis (FEA)

Formulating axisymmetric circular mesh stiffness matrices for high-stress aerospace pressure vessels.

Computer Graphics

Monte Carlo integration across circular lens apertures to generate physically accurate depth-of-field bokeh.

Fourier Optics

2D continuous Fourier transforms of circular pupil functions generating Airy disc diffraction intensity patterns.

Astrophysics

Integrating stellar surface brightness limb darkening profiles I(r) across circular planetary transits.

Foire Aux Questions sur l'Intégration de l'Aire du Cercle

Réponses aux questions les plus fréquentes sur les formules, le rayon, le diamètre et les unités

Comment prouver l'aire du cercle par intégration ?

Par couches concentriques : ∫₀ᴿ 2πr dr = πR². En coordonnées polaires : ∫₀²ᵖᶦ ∫₀ᴿ r dr dθ = πR².

Choisissez un outil de calcul spécialisé adapté à vos paramètres géométriques exacts

Main Tool

Area of Circle Calculator

Find circle area from radius, diameter, or circumference with instant formulas.

Outil de Calcul par Diamètre

Calculateur d'Aire de Cercle avec Diamètre

Calculate circle area directly from diameter using A = (πd²)/4.

Outil de Calcul par Circonférence

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Determine circle area directly from perimeter using A = C²/(4π).

Géométrie du Demi-Cercle

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Mécanique et Résistance des Matériaux

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