Verified Solution • 2025–26 Session moderate

QJEEM24Ma13ce: JEE Main 2024 Mathematics - Class 12 JEEM24Ma13ce Solution: The area of the region in the first quadrant inside the circle…

The area of the region in the first quadrant inside the circle x2+y2=8x^2+y^2=8 and outside the parabola y2=2xy^2=2 x is equal to :

Multiple Choice Options

  • A π2−13\frac{\pi}{2}-\frac{1}{3}
  • B π−13\pi-\frac{1}{3}
  • C π−23\pi-\frac{2}{3}
  • D π2−23\frac{\pi}{2}-\frac{2}{3}

Step-by-Step Verified Solution

We have, x2+y2=8x^2+y^2=8 and y2=2xy^2=2 x

JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Area Under The Curves Question 16 English Explanation

Area of shaded region

=∫02y22dy+∫2228−y2dy=16[y3]02+[y28−y2+4sin−1⁡(y22)]222=43+π−2=π−23

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