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JEE MainMathematicsArea Under Curves

The area of the region S = (x,y) : x^2 + y^2 32, y^2 4x is equal to

Options

  1. A8 3 (3 + 2)
  2. B8 3 (9 + 2)
  3. C8 3 (9 - 2)
  4. D4 3 (9 - 2)

Correct answer

C. 8 3 (9 - 2)

Step-by-step solution

The given region S is bounded by the circle x^2 + y^2 = 32 and the parabola y^2 = 4x . First, find the points of intersection of the two curves by substituting y^2 = 4x into the circle's equation: x^2 + 4x - 32 = 0 (x + 8)(x - 4) = 0 Since x 0 for real y on the parabola, x = 4 . This gives y = 4 . The inequality x^2 + y^2 32 represents the interior of the circle, and y^2 4x represents the region outside (to the left of) the parabola. The area of S is the total area of the circle minus the area of the smaller region

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