Area of Circle/Proof 6

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.

$\blacksquare$