Derivation of the Formula of a Circle

 

🎯 Goal

We want to derive the formula

$$A = \pi r^2$$

where $A$ is the area of the circle and $r$ is the radius.

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🧩 Step 1: Recall what a circle is

A circle is the set of all points that are at a fixed distance $r$ (the radius) from a central point.

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🧮 Step 2: Divide the circle into equal sectors

Imagine cutting the circle into many thin slices (like pizza slices).
If we rearrange these slices alternately (flipping every other one), they begin to resemble a parallelogram or rectangle shape.

As the number of slices increases, the shape more closely approximates a rectangle.

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📏 Step 3: Determine the dimensions of this "rectangle"

• Base (length): Half of the circumference of the circle

  $$\text{Base} = \frac{1}{2} \times \text{Circumference} = \frac{1}{2} \times 2\pi r = \pi r$$

• Height: Equal to the radius of the circle $r$

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📐 Step 4: Find the area of this “rectangle”

Area of rectangle = Base × Height

$$A = (\pi r) \times r = \pi r^2$$

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🌟 Step 5: The limiting argument (for precision)

As the number of sectors increases infinitely, the rearranged shape becomes exactly a rectangle — not just an approximation.
Thus, mathematically, the area of the circle is:

$$\boxed{A = \pi r^2}$$

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🔍 Step 6: Alternative Derivation (Using Integration)

For those familiar with calculus:

The area of a circle can also be found by integrating horizontal strips from $x = -r$ to $x = r$ under the curve:

$$x^2 + y^2 = r^2 \Rightarrow y = \sqrt{r^2 - x^2}$$

So,

$$A = 2 \int_{0}^{r} \sqrt{r^2 - x^2} \, dx$$

Using trigonometric substitution $x = r \sin\theta$, this becomes:

$$A = 2 \int_{0}^{\pi/2} r^2 \cos^2\theta \, d\theta = \pi r^2$$

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🌀 Step 7: Intuitive Meaning

• The π (pi) represents the constant ratio between a circle’s circumference and its diameter.

• The r² represents how the area scales — if you double the radius, the area quadruples.

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✅ Final Formula:

$$\boxed{A = \pi r^2}$$

Alternative derivation (step-by-step) using integration

We’ll compute the area of a circle of radius $r$ by integrating vertical slices.

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1. Equation of the circle and vertical slice

Circle: $x^2 + y^2 = r^2$.
Solve for the top half: $y = \sqrt{r^2 - x^2}$

For a given $x$ the vertical length of the circle is

$$\text{top} - \text{bottom} = \sqrt{r^2-x^2} - \big(-\sqrt{r^2-x^2}\big) = 2\sqrt{r^2-x^2}.$$

So the area is the integral of those vertical slices from $x=-r$ to $x=r$:

$$A=\int_{-r}^{r} 2\sqrt{r^2-x^2}\,dx .$$

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2. Use symmetry to simplify

The integrand is even, so

$$A = 2\int_{-r}^{r}\sqrt{r^2-x^2}\,dx = 4\int_{0}^{r}\sqrt{r^2-x^2}\,dx .$$

(We pulled a factor 2 outside and used $\int_{-r}^{r} = 2\int_{0}^{r}$.)

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3. Trigonometric substitution

Set $x = r\sin\theta$. Then

$$dx = r\cos\theta\,d\theta, \qquad \sqrt{r^2-x^2} = \sqrt{r^2 - r^2\sin^2\theta} = r\cos\theta.$$

Change the limits:

• $x=0 \Rightarrow \sin\theta=0 \Rightarrow \theta=0$,

• $x=r \Rightarrow \sin\theta=1 \Rightarrow \theta=\tfrac{\pi}{2}$.

Substitute into the integral:

$$\int_{0}^{r}\sqrt{r^2-x^2}\,dx = \int_{0}^{\pi/2} \big(r\cos\theta\big)\big(r\cos\theta\,d\theta\big) = r^2\int_{0}^{\pi/2}\cos^2\theta \,d\theta.$$

Thus

$$A = 4 \cdot r^2 \int_{0}^{\pi/2}\cos^2\theta \,d\theta.$$

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4. Evaluate $\displaystyle \int_{0}^{\pi/2}\cos^2\theta\,d\theta$

Use the identity $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$:

$$\int_{0}^{\pi/2}\cos^2\theta\,d\theta = \int_{0}^{\pi/2}\frac{1+\cos2\theta}{2}\,d\theta = \frac{1}{2}\int_{0}^{\pi/2}1\,d\theta \;+\; \frac{1}{2}\int_{0}^{\pi/2}\cos2\theta\,d\theta.$$

Compute each term:

• $\displaystyle \frac{1}{2}\int_{0}^{\pi/2}1\,d\theta = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4}$.

• $\displaystyle \frac{1}{2}\int_{0}^{\pi/2}\cos2\theta\,d\theta = \frac{1}{2}\left[\frac{\sin2\theta}{2}\right]_{0}^{\pi/2} = \frac{1}{4}\big(\sin\pi - \sin0\big) = 0.$

So

$$\int_{0}^{\pi/2}\cos^2\theta\,d\theta = \frac{\pi}{4}.$$

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5. Finish the computation

Plug that back into the area:

$$A = 4 r^2 \cdot \frac{\pi}{4} = \pi r^2.$$

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✅ Final result

$$\boxed{A=\pi r^2}$$

A=πr2​

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Short derivation using polar coordinates

Use polar coordinates $(\rho,\theta)$ where $x=\rho\cos\theta,\; y=\rho\sin\theta$.
The area element in polar coordinates is $dA=\rho\,d\rho\,d\theta$ (the Jacobian).

The disk of radius $r$ is described by $0\le \rho\le r,\; 0\le\theta\le 2\pi$. So

$$A=\iint_{D} dA=\int_{\theta=0}^{2\pi}\int_{\rho=0}^{r} \rho\,d\rho\,d\theta.$$

Evaluate the inner integral:

$$\int_{0}^{r}\rho\,d\rho=\frac{\rho^{2}}{2}\Big|_{0}^{r}=\frac{r^{2}}{2}.$$

Now the outer integral: