Let R be the smaller region bounded by the circle x^2 + y^2 = 8 and the parabola y^2 = 2x . Let T be the triangle formed by the origin and the two points of intersection of these curves. The area of the region that lies strictly inside R but outside the triangle T is equal to:

Step-by-step solution

First, find the points of intersection of the circle x^2 + y^2 = 8 and the parabola y^2 = 2x . Substituting y^2 = 2x into the circle's equation: x^2 + 2x - 8 = 0 (x + 4)(x - 2) = 0 Since x 0 for the parabola, x = 2 . For x = 2 , y^2 = 4 y = 2 . The points of intersection are P(2, 2) and Q(2, -2) . The triangle T is formed by the origin O(0,0) and the points P and Q . The base of the triangle along the vertical line x = 2 is PQ = 2 - (-2) = 4 . The height of the triangle from the origin to the line x = 2 is 2 . Area