JEE Main 202622 Jan, Evening shiftMathematicsEasyIntegral Calculus369× since 2002The area of the region A={(x,y):4x2+y2≤8 and y2≤4x}A=\{(x, y): 4x^2+y^2 \le 8 \text{ and } y^2 \le 4x\}A={(x,y):4x2+y2≤8 and y2≤4x} is :AOption A. π+4\pi+4π+4BOption B. π2+13\frac{\pi}{2}+\frac{1}{3}2π+31COption C. π+23\pi+\frac{2}{3}π+32DOption D. π2+2\frac{\pi}{2}+22π+2CheckSkip · show solution
