JEE Main 202227 Jul, Morning shiftMathematicsMediumIntegral Calculus369× since 2002The area of the smaller region enclosed by the curves y2=8x+4y^{2}=8 x+4y2=8x+4 and x2+y2+43x−4=0x^{2}+y^{2}+4 \sqrt{3} x-4=0x2+y2+43x−4=0 is equal toAOption A. 13(2−123+8π)\frac{1}{3}(2-12 \sqrt{3}+8 \pi)31(2−123+8π)BOption B. 13(2−123+6π)\frac{1}{3}(2-12 \sqrt{3}+6 \pi)31(2−123+6π)COption C. 13(4−123+8π)\frac{1}{3}(4-12 \sqrt{3}+8 \pi)31(4−123+8π)DOption D. 13(4−123+6π)\frac{1}{3}(4-12 \sqrt{3}+6 \pi)31(4−123+6π)CheckSkip · show solution
