Q.

Area of the region {(x,y):x2+(y-2)2≤4, x2≥2y} is        [2023]
 

1 2π-163      
2 π-83      
3 π+83      
4 2π+163  

Ans.

(1)

Given that x2+(y-2)2≤22 and x2≥2y

Convert the given inequalities into equations:

x2+(y-2)2=4 and x2=2y

Solving circle and parabola simultaneously:

2y+y2-4y+4=4, y2-2y=0, y=0, 2

Put y=2 in x2=2y⇒x=±2 ⇒(2,2) and (-2,2)

A1=2×2-14·π·22=4-π

Required area=2[∫02x22 dx-(4-π)]

=2[x36|02-4+π]=2[43+π-4]=2[π-83]=2π-163sq. units