Let the area of the region S = (x,y) : x^2+y^2 4, x^2 3|y| be . If 3 = + 3 , where and are integers, then the value of + is equal to

Step-by-step solution

The region S is the intersection of the interior of the circle x^2+y^2 = 4 and the region bounded by the parabolas x^2 = 3y (for y 0 ) and x^2 = -3y (for y Due to symmetry about both the x-axis and y-axis, the total area is 4 times the area in the first quadrant. In the first quadrant, the curves are x^2+y^2=4 and x^2=3y . Solving for their point of intersection: 3y + y^2 = 4 y^2 + 3y - 4 = 0 (y+4)(y-1) = 0 Since y 0 , we have y = 1 , which gives x = 3 . The area in the first quadrant is given by: A₁ = ₀^ 3 ( 4-x^2