nLab area enclosed by a circle

Contents

Context

Analysis

Geometry

Differential geometry

synthetic differential geometry

Introductions

from point-set topology to differentiable manifolds

geometry of physics: coordinate systems, smooth spaces, manifolds, smooth homotopy types, supergeometry

Differentials

V-manifolds

smooth space

Tangency

The magic algebraic facts

Theorems

Axiomatics

cohesion

tangent cohesion

differential cohesion

graded differential cohesion

singular cohesion

id id fermionic bosonic bosonic Rh rheonomic reduced infinitesimal infinitesimal & étale cohesive ʃ discrete discrete continuous * \array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& \esh &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }

Models

Lie theory, ∞-Lie theory

differential equations, variational calculus

Chern-Weil theory, ∞-Chern-Weil theory

Cartan geometry (super, higher)

Variational calculus

Contents

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 rr of a circle 𝒞\mathcal{C} in the Euclidean plane 2\mathbb{R}^2, the area of a circle is expressed either as A(r)=12τr 2A(r) = \frac{1}{2} \tau r^2 or as A(r)=πr 2A(r) = \pi r^2.

Proof by double integration

Proof

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

Given any Euclidean plane 2\mathbb{R}^2, one could select an orthonormal basis on 2\mathbb{R}^2 by postulating an origin 00 at the center of the circle 𝒞\mathcal{C} and two orthonormal vectors i^\hat{i} and j^\hat{j}. The circle 𝒞\mathcal{C} could be parameterized by a function r:[0,τ]×[0,r] 2\overrightarrow{r}:[0, \tau] \times [0, r] \to \mathbb{R}^2 defined as

r(ρ,θ)ρcos(θ)i^+ρsin(θ)j^\overrightarrow{r}(\rho, \theta) \coloneqq \rho \cos(\theta) \hat{i} + \rho \sin(\theta) \hat{j}

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

A(r)= 0 r 0 τ|r(ρ,θ)|dθdρA(r) = \int_{0}^{r} \int_{0}^{\tau} \vert \overrightarrow{r}(\rho, \theta) \vert d \theta d \rho

which evaluates to

A(r)= 0 r 0 τ|ρcos(θ)i^+ρsin(θ)j^|dθdρ= 0 r 0 τρ((cos(θ)) 2+(sin(θ)) 2)dθdρ= 0 r 0 τρdθdρ= 0 rτρdρ=12τrA(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 τ=2π\tau = 2 \pi.

Given any Euclidean plane 2\mathbb{R}^2, one could select an orthonormal basis on 2\mathbb{R}^2 by postulating an origin 00 at the center of the circle 𝒞\mathcal{C} and two orthonormal vectors i^\hat{i} and j^\hat{j}. There is an geometric algebra 𝔾 2\mathbb{G}^2 on the vector space defined by the equations i^ 2=1\hat{i}^2 = 1, j^ 2=1\hat{j}^2 = 1, and i^j^=j^i^\hat{i} \hat{j} = -\hat{j} \hat{i}.

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

r(θ)rcos(θ)i^+rsin(θ)j^\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}

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

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

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

A(r)= 0 τ|(rcos(θ)i^+rsin(θ)j^) θ(rcos(θ)i^+rsin(θ)j^)2|dθA(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)= 0 τ|(rcos(θ)i^+rsin(θ)j^)(rsin(θ)i^+rcos(θ)j^)2|dθA(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)= 0 τ|(r(cos(θ)) 2i^j^+r(sin(θ)) 2i^j^)2|dθA(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)= 0 τ|ri^j^2|dθ= 0 τr2dθ=12τrA(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 τ=2π\tau = 2 \pi.

Given any Euclidean plane 2\mathbb{R}^2, one could select an orthonormal basis on 2\mathbb{R}^2 by postulating an origin 00 at the center of the circle 𝒞\mathcal{C} and two orthonormal vectors i^\hat{i} and j^\hat{j}. The circle 𝒞\mathcal{C} could be parameterized by a function r:[0,τ] 2\overrightarrow{r}:[0, \tau] \to \mathbb{R}^2 defined as

r(θ)rcos(θ)i^+rsin(θ)j^\overrightarrow{r}(\theta) \coloneqq r \cos(\theta) \hat{i} + r \sin(\theta) \hat{j}

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

A(r)=12 0 τ|r(θ)| 2dθA(r) = \frac{1}{2} \int_{0}^{\tau} {\vert \overrightarrow{r}(\theta) \vert}^2 d \theta

which evaluates to

A(r)=12 0 τ|rcos(θ)i^+rsin(θ)j^| 2dθ=12 0 τ(r((cos(θ)) 2+(sin(θ)) 2)) 2dθ=12 0 τr 2dθ=12τrA(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 τ=2π\tau = 2 \pi.

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

A 𝒫(n)(r)=12r 2(2n)sin(τ2n)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 𝒫 : +()A_\mathcal{P}^\prime:\mathbb{R}_+ \to (\mathbb{R} \to \mathbb{R}), defined as

A 𝒫 (n)(r)=12r 2(2n)sin(τ2n)A_\mathcal{P}^\prime(n)(r) = \frac{1}{2} r^2 (2 n) \sin\left(\frac{\tau}{2n}\right)

The limit of A 𝒫 A_\mathcal{P}^\prime as nn goes to infinity is the area of a circle with radius rr:

A(r)=lim nA 𝒫 (n)(r)=lim n12r 2(2n)sin(τn)=12r 2lim m0sin(τm)m=12r 2lim m0 msin(τm) mm=12r 2lim m0τcos(τm)1=12τr 2A(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

Last revised on May 17, 2022 at 15:40:31. See the history of this page for a list of all contributions to it.