Question

Question #3fa8f

Answer 1

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`xx`area of one quadrant.
now, area of quadrant= `int_0^a(sqrt(a^2 - x^2)) dx` ...using(1)
`=[(x/a)(sqrt(a^2 - x^2)) +((a^2)/2)(sin^(-1)(x/a))]_"0"^a`
`=(pia^2)/4`
`rArr`area of circle is `4xx(pia^2)/4`
`=pia^2`

Answer 2

Consider a circle of radius `a`, We want to show that the area is `pia^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 \ 1/2 r^2 \ d theta`

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

Hence;

`A = int_0^(2pi) 1/2 a^2 \ d theta`
`\ \ \ = 1/2 \ a^2 \ int_0^(2pi) \ d theta`
`\ \ \ = 1/2 \ a^2 \ [theta]_0^(2pi)`
`\ \ \ = 1/2 \ a^2 \ (2pi-0)`
`\ \ \ = a^2 pi \ \` 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 `2pi`

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

`A = int int _R \ d A`

`\ \ \ = int_0^(2p) \ int_0^a \ r \ dr \ d theta`
`\ \ \ = int_0^(2pi) [ 1/2 r^2]_0^a \ d theta`
`\ \ \ = int_0^(2pi) ( 1/2 a^2 - 0) \ d theta`
`\ \ \ = 1/2 a^2 \ int_0^(2pi) \ d theta`
`\ \ \ = 1/2 a^2 \ [ theta]_0^(2pi) \ d theta`
`\ \ \ = 1/2 a^2 \ (2pi-0)`
`\ \ \ = a^2pi \ \` QED