JEE Main 20178 Apr, Morning shiftMathematicsHardIntegral Calculus369× since 2002The area (in sq. units) of the smaller portion enclosed between the curves, x² + y² = 4 and y² = 3x, is :AOption A. 123+π3{1 \over {2\sqrt 3 }} + {\pi \over 3}231+3πBOption B. 13+2π3{1 \over {\sqrt 3 }} + {{2\pi } \over 3}31+32πCOption C. 123+2π3{1 \over {2\sqrt 3 }} + {{2\pi } \over 3}231+32πDOption D. 13+4π3{1 \over {\sqrt 3 }} + {{4\pi } \over 3}31+34πCheckSkip · show solution
