Question

Area lying in the first quadrant and bounded by the circle x 2 + y 2 = 4 and the lines x = 0 and x = 2 is A. π B. C. D.

Answer 1

We have to find the area in the first quadrant, bounded by the circle x 2 + y 2 =4 and lines x=0 and x=2. Draw the graph of these equations.

Figure (1)

To calculate the area of the region AOBA, we take a vertical strip in the region with infinitely small width as shown in the figure above.

To find the area of the region AOBA, integrate the area of the strip.

Area of the region AOBA= ∫ 0 2 ydx (1)

The equation of the circle is x 2 + y 2 =4. From this equation find the value of y in terms of x and substitute in equation (1).

x 2 + y 2 =4 y 2 =4− x 2 y= 4− x 2

Substitute 4− x 2 for y in equation (1).

Area of the region AOBA= ∫ 0 2 4− x 2 dx = [ x 2 ( ( 2 ) 2 − x 2 )+ ( 2 ) 2 2 sin −1 x 2 ] 0 2 =[ 2 2 ( ( 2 ) 2 − 2 2 )+ ( 2 ) 2 2 sin −1 2 2 −( 0 2 ( ( 2 ) 2 − ( 0 ) 2 )+ ( 2 ) 2 2 sin −1 0 2 ) ] =π sq units

The area calculated for the region AOBA is π sq units.

Thus, out of all the four options, option (A) is correct.