JEE Main 20248 Apr, Evening shiftMathematicsMediumIntegral Calculus369× since 2002The area of the region in the first quadrant inside the circle x2+y2=8x^2+y^2=8x2+y2=8 and outside the parabola y2=2xy^2=2 xy2=2x is equal to :AOption A. π2−13\frac{\pi}{2}-\frac{1}{3}2π−31BOption B. π−13\pi-\frac{1}{3}π−31COption C. π−23\pi-\frac{2}{3}π−32DOption D. π2−23\frac{\pi}{2}-\frac{2}{3}2π−32CheckSkip · show solution
