nLab area enclosed by a circle

Idea

The area enclosed by a circle as discussed in Euclidean geometry.

Definition/proposition and proofs

Strictly speaking, we are talking about the area of the disk whose boundary is the circle; however, the average person usually identifies the interior of a geometric shape with its boundary.

Proposition

Depending on which circle constant you use, given a radius $r$ of a circle $\mathcal{C}$ in the Euclidean plane $\mathbb{R}^2$, the area of a circle is expressed either as $A(r) = \frac{1}{2} \tau r^2$ or as $A(r) = \pi r^2$.

Proof by double integration

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

Given any Euclidean plane $\mathbb{R}^2$, one could select an orthonormal basis on $\mathbb{R}^2$ by postulating an origin $0$ at the center of the circle $\mathcal{C}$ and two orthonormal vectors $\hat{i}$ and $\hat{j}$. The circle $\mathcal{C}$ could be parameterized by a function $\overrightarrow{r}:[0, \tau] \times [0, r] \to \mathbb{R}^2$ defined as

$$\overrightarrow{r}(\rho, \theta) \coloneqq \rho \cos(\theta) \hat{i} + \rho \sin(\theta) \hat{j}$$

Then the area of $\mathcal{C}$ is given by the following double integral?:

$$A(r) = \int_{0}^{r} \int_{0}^{\tau} \vert \overrightarrow{r}(\rho, \theta) \vert d \theta d \rho$$

which evaluates to

$$A(r) = \int_{0}^{r} \int_{0}^{\tau} \vert \rho \cos(\theta) \hat{i} + \rho \sin(\theta) \hat{j} \vert d \theta d \rho = \int_{0}^{r} \int_{0}^{\tau} \rho((\cos(\theta))^2 + (\sin(\theta))^2) d \theta d \rho = \int_{0}^{r} \int_{0}^{\tau} \rho d \theta d \rho = \int_{0}^{r} \tau \rho d \rho = \frac{1}{2} \tau r$$

Proof by areal velocity

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

Given any Euclidean plane $\mathbb{R}^2$, one could select an orthonormal basis on $\mathbb{R}^2$ by postulating an origin $0$ at the center of the circle $\mathcal{C}$ and two orthonormal vectors $\hat{i}$ and $\hat{j}$. There is an geometric algebra $\mathbb{G}^2$ on the vector space defined by the equations $\hat{i}^2 = 1$, $\hat{j}^2 = 1$, and $\hat{i} \hat{j} = -\hat{j} \hat{i}$.

The circle $\mathcal{C}$ could be parameterized by a function $\overrightarrow{r}:[0, \tau] \to \mathbb{R}^2$ defined as

$$\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}$$

Then the area of $\mathcal{C}$ is given by integrating the magnitude of the areal velocity:

$$A(r) = \int_{0}^{\tau} \left|\frac{\overrightarrow{r}(\theta) \wedge \overrightarrow{v}(\theta)}{2}\right| d \theta$$

where $a \wedge b$ is the wedge product of two multivectors $a$ and $b$ and $\overrightarrow{v}$ is the velocity of a point in $\mathcal{C}$. This expression evaluates to

$$A(r) = \int_{0}^{\tau} \left|\frac{(r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}) \wedge \partial_\theta (r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j})}{2}\right| d \theta$$

$$A(r) = \int_{0}^{\tau} \left|\frac{(r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}) \wedge (-r \sin(\theta) \hat{i} + r \cos(\theta) \hat{j})}{2}\right| d \theta$$

$$A(r) = \int_{0}^{\tau} \left|\frac{(r (\cos(\theta))^2 \hat{i} \hat{j} + r (\sin(\theta))^2 \hat{i} \hat{j})}{2}\right| d \theta$$

$$A(r) = \int_{0}^{\tau} \left|\frac{r \hat{i} \hat{j}}{2}\right| d \theta = \int_{0}^{\tau} \frac{r}{2} d \theta = \frac{1}{2} \tau r$$

Proof by action functionals

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

Given any Euclidean plane $\mathbb{R}^2$, one could select an orthonormal basis on $\mathbb{R}^2$ by postulating an origin $0$ at the center of the circle $\mathcal{C}$ and two orthonormal vectors $\hat{i}$ and $\hat{j}$. The circle $\mathcal{C}$ could be parameterized by a function $\overrightarrow{r}:[0, \tau] \to \mathbb{R}^2$ defined as

$$\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}$$

Then the area of $\mathcal{C}$ is given by the action functional of the parameterized curve:

$$A(r) = \frac{1}{2} \int_{0}^{\tau} {\vert \overrightarrow{r}(\theta) \vert}^2 d \theta$$

which evaluates to

$$A(r) = \frac{1}{2} \int_{0}^{\tau} {\vert r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j} \vert}^2 d \theta = \frac{1}{2} \int_{0}^{\tau} (r((\cos(\theta))^2 + (\sin(\theta))^2))^2 d \theta = \frac{1}{2} \int_{0}^{\tau} r^2 d \theta = \frac{1}{2} \tau r$$

Proof by limits of regular polygons

Proof

In this proof, we are using the circle constant $\tau = 2 \pi$.

The area of a regular polygon $\mathcal{P}_n$ with $n$ sides and circumradius $r$ is given by the sequence of functions $P:\mathbb{N} \to (\mathbb{R} \to \mathbb{R})$

$$A_\mathcal{P}(n)(r) = \frac{1}{2} r^2 (2 n) \sin\left(\frac{\tau}{2 n}\right)$$

which embeds in the $\mathbb{R}_+$-action $A_\mathcal{P}^\prime:\mathbb{R}_+ \to (\mathbb{R} \to \mathbb{R})$, defined as

$$A_\mathcal{P}^\prime(n)(r) = \frac{1}{2} r^2 (2 n) \sin\left(\frac{\tau}{2n}\right)$$

The limit of $A_\mathcal{P}^\prime$ as $n$ goes to infinity is the area of a circle with radius $r$:

$$A(r) = \lim_{n \to \infty} A_\mathcal{P}^\prime(n)(r) = \lim_{n \to \infty} \frac{1}{2} r^2 (2 n) \sin\left(\frac{\tau}{n}\right) = \frac{1}{2} r^2 \lim_{m \to 0} \frac{\sin(\tau m)}{m} = \frac{1}{2} r^2 \lim_{m \to 0} \frac{\partial_m \sin(\tau m)}{\partial_m m} = \frac{1}{2} r^2 \lim_{m \to 0} \frac{\tau \cos(\tau m)}{1} = \frac{1}{2} \tau r^2$$

See also

• circle

• unit circle

• area

• circumference of a circle

• Euclidean space