Smaller area enclosed by the circle \(x^2+y^2=16\) exterior to the parabola \(y^2=6x\) is1\(\frac{4}{3}(4\pi-\sqrt{3})\)2\(\frac{4}{3}(4\pi+\sqrt{3})\)3\(\frac{4}{3}(8\pi-\sqrt{3})\)4\(\frac{4}{3}(8\pi+\sqrt{3})\)Answer & explanation+Correct answer: option 2Use integration
