# Question #3fa8f

Mar 26, 2017

see explanation

#### Explanation:

consider circle ${x}^{2} + {y}^{2} = {a}^{2}$ , (a is the radius) ...(1).
divide the circle in four quadrants, so area of circle=4$\times$area of one quadrant.
now, area of quadrant= ${\int}_{0}^{a} \left(\sqrt{{a}^{2} - {x}^{2}}\right) \mathrm{dx}$ ...using(1)
$= {\left[\left(\frac{x}{a}\right) \left(\sqrt{{a}^{2} - {x}^{2}}\right) + \left(\frac{{a}^{2}}{2}\right) \left({\sin}^{- 1} \left(\frac{x}{a}\right)\right)\right]}_{\text{0}}^{a}$
$= \frac{\pi {a}^{2}}{4}$
$\Rightarrow$area of circle is $4 \times \frac{\pi {a}^{2}}{4}$
$= \pi {a}^{2}$

Mar 26, 2017

Consider a circle of radius $a$, We want to show that the area is $\pi {a}^{2}$

Method 1 - Polar Area Formula

We will calculate the area using Polar Coordinates. Polar area is given by the formula;

$A = {\int}_{\alpha}^{\beta} \setminus \frac{1}{2} {r}^{2} \setminus d \theta$

For our circle of radius $a$, we have $r = a$ (fixed constant) and $\theta \in \left[0 , 2 \pi\right]$,

Hence;

$A = {\int}_{0}^{2 \pi} \frac{1}{2} {a}^{2} \setminus d \theta$
$\setminus \setminus \setminus = \frac{1}{2} \setminus {a}^{2} \setminus {\int}_{0}^{2 \pi} \setminus d \theta$
$\setminus \setminus \setminus = \frac{1}{2} \setminus {a}^{2} \setminus {\left[\theta\right]}_{0}^{2 \pi}$
$\setminus \setminus \setminus = \frac{1}{2} \setminus {a}^{2} \setminus \left(2 \pi - 0\right)$
$\setminus \setminus \setminus = {a}^{2} \pi \setminus \setminus$ QED

Method 2 - Double Integral

We could also consider a double integral, Suppose $R$ is the area bounded by the circle; then:

$r$ would be a ray varying from $0$ to $a$;
$\theta$ would vary from $0$ to $2 \pi$

The area, $A$, of the circle would then be given by:

$A = \int {\int}_{R} \setminus d A$

$\setminus \setminus \setminus = {\int}_{0}^{2 p} \setminus {\int}_{0}^{a} \setminus r \setminus \mathrm{dr} \setminus d \theta$
$\setminus \setminus \setminus = {\int}_{0}^{2 \pi} {\left[\frac{1}{2} {r}^{2}\right]}_{0}^{a} \setminus d \theta$
$\setminus \setminus \setminus = {\int}_{0}^{2 \pi} \left(\frac{1}{2} {a}^{2} - 0\right) \setminus d \theta$
$\setminus \setminus \setminus = \frac{1}{2} {a}^{2} \setminus {\int}_{0}^{2 \pi} \setminus d \theta$
$\setminus \setminus \setminus = \frac{1}{2} {a}^{2} \setminus {\left[\theta\right]}_{0}^{2 \pi} \setminus d \theta$
$\setminus \setminus \setminus = \frac{1}{2} {a}^{2} \setminus \left(2 \pi - 0\right)$
$\setminus \setminus \setminus = {a}^{2} \pi \setminus \setminus$ QED