6. Smaller area enclosed by the circle x2+ y2 4 and the line xy 2 is(D) 2 (π + 2)(A) 2 (1-2)(B) π-2(C) 21
We have to find the smaller area enclosed by the circle whose equation is x 2 + y 2 = 4 , and the line x + y = 2 .
Plot the equations and shade the smaller region bounded by the circle and line.
Figure (1)
The area of the region ABCA is to be calculated.
The area of the region OBCAO lies in the first quadrant.
To find the area of the region OBCAO, assume a vertical strip of infinitely small width and integrate it.
Area of the region OBCAO = ∫ 0 2 y d x = ∫ 0 2 4 − x 2 d x
Simplify further,
Area of the region OBCAO = [ x 2 2 2 − x 2 + 2 2 2 sin − 1 x 2 ] 0 2 = [ 2 ( π 2 ) ] = π
Similarly find the area bounded by the straight line x + y = 2 with the x-axis.
Area of the triangle OAB = ∫ 0 2 ( 2 − x ) d x = [ 2 x − x 2 2 ] 0 2 = [ 2 ( 2 ) − ( 2 ) 2 2 − ( 2 ( 0 ) − ( 0 ) 2 2 ) ] = 2 sq units
The area of the shaded region is,
Area of the region ABCA = Area of the region OBCAO − Area of the triangle OAB = ( π − 2 ) sq units
The area of the triangular region is ( π − 2 ) sq units .
Thus, out of all the four options, option (B) is correct.
We have to find the smaller area enclosed by the circle whose equation is
Plot the equations and shade the smaller region bounded by the circle and line.
Figure (1)
The area of the region ABCA is to be calculated.
The area of the region OBCAO lies in the first quadrant.
To find the area of the region OBCAO, assume a vertical strip of infinitely small width and integrate it.
Simplify further,
Similarly find the area bounded by the straight line
The area of the shaded region is,
The area of the triangular region is
Thus, out of all the four options, option (B) is correct.
