Using integration, find the area of the region enclosed between the two circles `x^2+y^2=4` and `(x-2)^2+y^2=4.`

Explanation

Equations of the given circles are `x^2+y^2=4 "(eq 1)" and (x-2)^2+y^2-4 "(eq 1)" `
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C (2, 0) and radius 2.
Solving equations (1) and (2), we have
`x=1 and y=+-sqrt3` Thus, the points of intersection of the given circles are A (1,3) and A′ (1,–3)

Required area of the enclosed region OACA′O between circles= 2 [area of the region ODCAO]
= 2 [area of the region ODAO + area of the region DCAD]
`2 [int_0^2 ydx +int_1^2 y dx]`
`2[int_0^1(sqrt(4-(x-2)^2)dx +int_1^2 sqrt(4-x^2)dx]`
now by using formula `intsqrt(x^2-x^2)=x/2 sqrt(x^2-a^2)+a^2/2sin^-1(x/a)+C`

by solving above integration and putting the value of the limits we will get
`(-sqrt3-(2pi)/3+2 pi)+ (2 pi -sqrt3 -(2 pi)/3)`
so by solving and simplifying this we will get `(8pi)/3-2sqrt3` as a final answer