Area of Circle

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Theorem

The area $A$ of a circle is given by:

$A = \pi r^2$

where $r$ is the radius of the circle.


Proof 1

From Equation of Circle:

$x^2 + y^2 = r^2$

Thus $y = \pm \sqrt {r^2 - x^2}$.


It follows that from the geometric interpretation of the definite integral:

\(\ds A\) \(=\) \(\ds \int_{-r}^r \paren {\sqrt {r^2 - x^2} - \paren {-\sqrt {r^2 - x^2} } } \rd x\)
\(\ds \) \(=\) \(\ds \int_{-r}^r 2 \sqrt {r^2 - x^2} \rd x\)
\(\ds \) \(=\) \(\ds \int_{-r}^r 2 r \sqrt {1 - \frac {x^2} {r^2} } \rd x\)

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

Thus $\theta = \map \arcsin {\dfrac x r}$ and $\rd x = r \cos \theta \rd \theta$.

\(\ds A\) \(=\) \(\ds \int_{\map \arcsin {\frac {-r} r} }^{\map \arcsin {\frac r r} } 2 r^2 \sqrt {1 - \frac {\paren {r \sin \theta}^2} {r^2} } \cos \theta \rd \theta\) Integration by Substitution
\(\ds \) \(=\) \(\ds \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta\)
\(\ds \) \(=\) \(\ds \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {\cos^2 \theta} \cos \theta \rd \theta\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds r^2 \int_{-\frac \pi 2}^{\frac \pi 2} 2 \cos^2 \theta \rd \theta\)
\(\ds \) \(=\) \(\ds r^2 \int_{-\frac \pi 2}^{\frac \pi 2} \paren {1 + \map \cos {2 \theta} } \rd \theta\) Double Angle Formula for Cosine: Corollary 1
\(\ds \) \(=\) \(\ds r^2 \intlimits {\theta + \frac 1 2 \map \sin {2 \theta} } {-\frac \pi 2} {\frac \pi 2}\) from Integration of Constant and Primitive of Cosine Function
\(\ds \) \(=\) \(\ds r^2 \paren {\frac \pi 2 + \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac \pi 2} }\)
\(\ds \) \(=\) \(\ds r^2 \paren {2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}\)
\(\ds \) \(=\) \(\ds \pi r^2\)

$\blacksquare$


Proof 2

Proof by shell integration:

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


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\(\ds A\) \(=\) \(\ds \int_0^r 2 \pi t \rd t\) Perimeter of Circle
\(\ds \) \(=\) \(\ds \bigintlimits {\pi t^2} 0 r\)
\(\ds \) \(=\) \(\ds \pi r^2\)

$\blacksquare$


Proof 3

Area-of-Circle-Proof-3.png

Construct a circle with radius $r$ and circumference $c$, whose area is denoted by $C$.

Construct a triangle with height $r$ and base $c$, whose area is denoted by $T$.


Lemma $1$

$T = \pi r^2$

$\Box$


Lemma $2$

$T \ge C$

$\Box$


Lemma $3$

$T \le C$

$\Box$


Final Proof

From Lemma $2$:

$T \ge C$

From Lemma $3$:

$T \le C$

Therefore:

$T \mathop = C$

and so from Lemma $1$:

$C \mathop = T \mathop = \pi r^2$

$\blacksquare$


Proof 4

Expressing the area in polar coordinates:

\(\ds \iint \rd A\) \(=\) \(\ds \int_0^r \int_0^{2 \pi} t \rd t \rd \theta\)
\(\ds \) \(=\) \(\ds \intlimits {\int_0^r t \theta} 0 {2 \pi} \rd t\)
\(\ds \) \(=\) \(\ds \int_0^r 2 \pi t \rd t\)
\(\ds \) \(=\) \(\ds 2 \pi \paren {\intlimits {\frac 1 2 t^2} 0 r}\)
\(\ds \) \(=\) \(\ds 2 \pi \paren {\frac 1 2 r^2}\)
\(\ds \) \(=\) \(\ds \pi r^2\)

$\blacksquare$


Kepler's Proof

Let the circle of radius $r$ be divided into many sectors:

Area-of-Circle-Kepler's-Proof.png

If they are made small enough, they can be approximated to triangles whose heights are all $r$.

Let the bases of these triangles be denoted:

$b_1, b_2, b_3, \ldots$

From Area of Triangle in Terms of Side and Altitude, their areas are:

$\dfrac {r b_1} 2, \dfrac {r b_2} 2, \dfrac {r b_3} 2, \ldots$

The area $\AA$ of the circle is given by the sum of the areas of each of these triangles:

\(\ds \AA\) \(=\) \(\ds \dfrac {r b_1} 2 + \dfrac {r b_2} 2 + \dfrac {r b_3} 2 + \cdots\)
\(\ds \) \(=\) \(\ds \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}\)

But $b_1 + b_2 + b_3 + \cdots$ is the length of the circumference of the circle.

From Perimeter of Circle:

$b_1 + b_2 + b_3 + \cdots = 2 \pi r$

Hence:

\(\ds \AA\) \(=\) \(\ds \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}\)
\(\ds \) \(=\) \(\ds \dfrac r 2 \paren {2 \pi r}\)
\(\ds \) \(=\) \(\ds \pi r^2\)

It needs to be noted that this proof is intuitive and non-rigorous.

$\blacksquare$


Proof 6

From Equation of Circle:

$x^2 + y^2 = r^2$


Let $A$ be the area of the circle whose equation is given by $x^2 + y^2 = r^2$.

We have that:

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

For the upper half of the circle:

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

Thus for the right hand half of the upper half of the circle:

\(\ds \frac A 4\) \(=\) \(\ds \int_0^r \sqrt {r^2 - x^2} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\pi r^2} 4\) Definite Integral from $0$ to $r$ of $\sqrt {r^2 - x^2}$

Hence the result.

$\blacksquare$


Proof 7

By the method of exhaustion:

AreaOfCircleMethodOfExhaustion.png


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Sources

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