Area of a Circle

Theorem

The area $A$ of a circle is given by the formula $A=\pi r^2$, where $r$ is the radius of the circle.

Proof

We start with the equation of a circle: $x^2 + y^2 = r^2$.

Thus $y = \pm \sqrt{r^2 - x^2}$, so from the geometric interpretation of the definite integral:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle A\)	\(=\)	\(\displaystyle \int_{-r}^r \left[ \sqrt{r^2 - x^2} - (-\sqrt{r^2 - x^2})\right] \mathrm d x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_{-r}^r 2 \sqrt{r^2 - x^2} \ \mathrm d x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_{-r}^r 2 r \sqrt{1 - \frac{x^2}{r^2} } \ \mathrm d x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Let $x = r \sin \theta$ (note that we can do this because $-r \le x \le r$).

Thus $\theta = \arcsin \left({\dfrac x r}\right)$ and $\mathrm d x = r \cos \theta \ \mathrm d \theta$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle A\)	\(=\)	\(\displaystyle \int_{\arcsin(\frac{-r} r)}^{\arcsin(\frac r r)} 2r^2 \sqrt{1-\frac{(r \sin \theta)^2}{r^2} }\cos \theta \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Integration by Substitution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_{-\frac{\pi} 2}^{\frac{\pi} 2} 2r^2\sqrt{1-\sin^2\theta}\cos\theta \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \int_{-\frac{\pi} 2}^{\frac{\pi} 2} 2r^2\sqrt{\cos^2\theta}\cos\theta \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Pythagorean trigonometric identities
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle r^2\int_{-\frac{\pi} 2}^{\frac{\pi} 2} 2\cos^2\theta \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle r^2\int_{-\frac{\pi} 2}^{\frac{\pi} 2} (1+\cos(2\theta)) \ \mathrm d \theta\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Since $2\cos^2\theta = 1 + \cos(2\theta)$ from the double angle formula for cosine
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle r^2\left[\theta + \frac 1 2 \sin(2\theta)\right]_{-\frac{\pi} 2}^{\frac{\pi} 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From Integration of a Constant and Integral of Cosine Function
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle r^2\left[\frac{\pi} 2 + \frac 1 2 \sin\left(2\cdot\frac{-\pi} 2\right) - \frac{-\pi} 2 - \frac 1 2 \sin \left(2 \cdot \frac {\pi} 2 \right)\right]\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle r^2\left[2\cdot\frac{\pi} 2 + 2\cdot\frac 1 2 \cdot 0 \right]\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pi r^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof by Shell Integration

The circle can be divided into a set of infinitesimally thin rings, each of which has area $2 \pi t \ \mathrm dt$, since the ring has length $2 \pi t$ and thickness $\mathrm d t$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle A\)	\(=\)	\(\displaystyle \int_0^r 2\pi t \ \mathrm d t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	From the definition of pi
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left[{\pi t^2}\right]_0^r\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \pi r^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $4.11$