Area of Circle/Proof 1

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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

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 Definite Integral of Constant and Primitive of Cosine Function
\(\ds \) \(=\) \(\ds r^2 \paren {\frac \pi 2 + \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac \pi 2} }\)
\(\ds \) \(=\) \(\ds r^2 \paren {2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}\)
\(\ds \) \(=\) \(\ds \pi r^2\)

$\blacksquare$


Sources

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