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JEE MainMathematicsArea Under Curves

The area of the region S = (x,y) : x^2 + y^2 16, x^2 6y, x 0 is equal to :

Options

  1. A8 3 + 2 3 3
  2. B4 3 - 2 3 3
  3. C8 3 - 2 3 3
  4. D4 3 + 2 3 3

Correct answer

A. 8 3 + 2 3 3

Step-by-step solution

The region S is bounded by the circle x^2 + y^2 = 16 , the parabola x^2 = 6y , and the y-axis ( x 0 ). First, find the point of intersection of the circle and the parabola in the first quadrant. Substituting x^2 = 6y into the circle's equation: 6y + y^2 = 16 y^2 + 6y - 16 = 0 (y+8)(y-2) = 0 Since y 0 , we have y = 2 . The corresponding x-coordinate is x = 6(2) = 2 3 . The region S is bounded above by the circle y = 16-x^2 and below by the parabola y = x^2 6 for x [0, 2 3 ] . The required area is given by: A = ₀^ 2

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