Area of Circle/Proof 6
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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:
- $x^2 + y^2 = r^2$
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.
$\blacksquare$