The area of the smaller region enclosed by the curves y 2 = 8 x + 4 and x 2 + y 2 + 4 3 x - 4 = 0 is equal to

Step-by-step solution

Plotting the area of the smaller region enclosed by the curves y 2 = 8 x + 4 and x 2 + y 2 + 4 3 x - 4 = 0 we get, Now finding point of intersection of x 2 + y 2 + 4 3 x - 4 = 0 and y 2 = 8 x + 4 We get, 0 , 2 and 0 , - 2 Both are symmetric about x - axis So, area will be = 2 ∫ 0 2 16 - y 2 - 2 3 - y 2 - 4 8 d y = 2 1 2 y 16 - y 2 + 16 sin - 1 y 4 - 2 3 y - y 3 24 + 1 2 y 0 2 = 1 3 8 π + 4 - 12 3