Question #3fa8f

2 Answers
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=4xxarea 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

Mar 26, 2017

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