Determine the area of the region ( x , y ) ∈ ℝ 2 : x 2 + y 2 ≤ a 2 , x ≥ a 2 , a > 0 .

Given x 2 + y 2 ≤ a 2 , x ≥ a 2 , a > 0 Consider x 2 + y 2 = a 2 , x = a 2 , a > 0 Area of the bounded region is A = 2 ∫ a 2 a a 2 - x 2 d x = 2 1 2 x a 2 - x 2 + 1 2 a 2 sin - 1 x a a 2 a = 2 1 2 a a 2 - a 2 + 1 2 a 2 sin - 1 a a - 2 1 2 a 2 a 2 - a 4 2 + 1 2 a 2 sin - 1 a 2 a = 2 0 + 1 2 a 2 sin - 1 1 - 2 1 2 a 2 3 a 2 + 1 2 a 2 sin - 1 1 2 = 2 a 2 2 π 2 - 2 3 a 2 8 + a 2 2 π 6 = 2 πa 2 4 - 2 3 a 2 8 + πa 2 16 = a 2 π 3 - 3 4 sq.units.

Find the area of the region ( x , y ) ∈ ℝ 2 : x 2 + y 2 ≤ 4 , x + y ≥ 2

Let R = { ( x , y ) : x 2 + y 2 ≤ 4 , x + y ≥ 2 } ⇒ R = { ( x , y ) : x 2 + y 2 ≤ 4 } ∩ { ( x , y ) : x + y ≥ 2 } i.e., R = R 1 ∩ R 2 , where R 1 = { ( x , y ) : x 2 + y 2 ≤ 4 } and R 2 = { ( x , y ) : x + y ≥ 2 } For region R 1 : x 2 + y 2 = 4 is a circle having centre at ( 0 , 0 ) and radius 2 . Since ( 0 , 0 ) satisfy the equation x 2 + y 2 ≤ 4 , region R 1 is the region interior of the circle x 2 + y 2 = 4 For region R 2 : x + y = 2 is a straight line passing through ( 0 , 2 ) and ( 2 , 0 ) which divides the XY-plane in two parts. Since ( 0 , 0 ) does not satisfy x + y ≥ 2 , R 2 is that region which does not contain origin ( 0 , 0 ) ie., above the line x + y = 2 . So, the shaded region is the required region ∴ Area of the required region = ∫ 0 2 4 − x 2 d x − ∫ 0 2 ( 2 − x ) d x = [ 1 2 x 4 − x 2 + 1 2 4 sin − 1 ( x 2 ) ] 2 0 − 2 [ x ] 2 0 + [ x 2 2 ] 2 0 = [ 2 sin − 1 1 − 0 ] − 2 [ 2 − 0 ] + 4 2 − 0 = 2 × π 2 − 4 + 2 = ( π − 2 ) sq . units .

Obtain the area of the region ( x , y ) ∈ ℝ 2 : x 2 + y 2 ≤ 32 , x ≥ y ≥ 0

Given x 2 + y 2 ≤ 32 , x ≥ y ≥ 0 Consider x 2 + y 2 = 32 . . . 1 and y = x and y = 0 When y = x then x = ± 4 When y = 0 then x = ± 4 2 Area of the bounded region = ∫ 0 4 x d x + ∫ 4 4 2 32 - x 2 d x = x 2 2 0 4 + 1 2 x 32 - x 2 + 16 sin - 1 x 4 2 4 4 2 = 8 + 16 π 2 - 8 - 16 π 4 = 4 π sq . unit

Find the area of the region ( x , y ) ∈ ℝ 2 : x 2 + ( y + 2 ) 2 ≤ 16 , y ≥ 0

Given x 2 + y + 2 2 = 16 , this represents equation of circle whose centre is at 0 , - 2 & radius is 4 . Sketch Curve is symmetric about y - a x i s ∴ Required Area = 2 ∫ 0 2 3 - 2 + 16 - x 2 d x = 2 - 2 x + x 2 16 - x 2 + 16 2 sin - 1 x 4 0 2 3 = 2 - 4 3 + 2 3 2 16 - 12 + 8 sin - 1 2 3 4 - 0 = 2 - 4 3 + 2 3 + 8 sin - 1 3 2 = 2 - 2 3 + 8 × π 3 = 16 π 3 - 4 3 s q . u n i t