Question

Find the area bounded by curves ( x – 1) 2 + y 2 = 1 and x 2 + y 2 = 1

Answer 1

We have to find the area bounded by curves ( x−1 ) 2 + y 2 =1 and x 2 + y 2 =1. Draw the graphs of the two equations and shade the common region.

Figure (1)

The shaded region AOBCA is to be calculated. From the above figure, it can be observed that the shaded region is symmetric about x-axis. So, the area of AOBCA is twice the area of OACO.

The given equations are,

x 2 + y 2 =1(1)

( x−1 ) 2 + y 2 =1(2)

Substitute 1− x 2 for y 2 in second equation of circle.

( x−1 ) 2 +1− x 2 =1 x 2 +1−2x− x 2 =0 x= 1 2

The y coordinate of points A and B are,

x 2 + y 2 =1 ( 1 2 ) 2 + y 2 =1 y 2 =1− 1 4 y=± 3 2

Draw a perpendicular from point A to B intersecting x-axis at M. The coordinates of point M are ( 1 2 ,0 ).

The area of OCAO is the sum of areas of OMAO and MCAM.

The area of OMAO is calculated as,

Area of the region OMAO= ∫ 0 1 2 ydx = ∫ 0 1 2 1−( x−1 ) dx

The area of OMBO is calculated as,

Area of the region MCAM= ∫ 1 2 1 ydx = ∫ 1 2 1 1− x 2 dx

The equation for the area of OCAO is,

Area of OCAO=Area of OMAO+Area of MCAM = ∫ 0 1 2 1−( x−1 ) dx + ∫ 1 2 1 1− x 2 dx

Simplify further,

Area of the region OCAO= [ x−1 2 1− ( x−1 ) 2 + 1 2 sin −1 ( x−1 ) ] 0 1 2 + [ x 2 1− ( x ) 2 + 1 2 sin −1 x ] 1 2 1 ={ [ 1 2 −1 2 1− ( 1 2 −1 ) 2 + 1 2 sin −1 ( 1 2 −1 )−( 0−1 2 1− ( 0−1 ) 2 + 1 2 sin −1 ( 0−1 ) ) ] +[ 1 2 1− ( 1 ) 2 + 1 2 sin −1 1−( 1 2 2 1− ( 1 2 ) 2 + 1 2 sin −1 1 2 ) ] } =[ − 3 8 + 1 2 ( − π 6 )− 1 2 ( − π 2 ) ]+[ 1 2 ( π 2 )− 3 8 − 1 2 ( π 6 ) ]

Simplify further,

Area of the region OCAO=[ − 3 4 − π 12 + π 4 + π 4 − π 12 ] =[ − 3 4 − π 6 + π 2 ] =[ 2π 6 − 3 4 ]

The area of OBCAO is twice the area of OCAO,

Area of the region OBCAO=2×Area of the region OCAO =2×( 2π 6 − 3 4 ) = 2π 3 − 3 2  sq units

Thus, the required area bounded by the two circles is ( 2π 3 − 3 2 ) sq units.