CUET Preparation Today
| Smaller area enclosed by the circle \(x^2+y^2=16\) exterior to the parabola \(y^2=6x\) is |
\(\frac{4}{3}(4\pi-\sqrt{3})\) \(\frac{4}{3}(4\pi+\sqrt{3})\) \(\frac{4}{3}(8\pi-\sqrt{3})\) \(\frac{4}{3}(8\pi+\sqrt{3})\) |
| \(\frac{4}{3}(4\pi+\sqrt{3})\) |
| Use integration |
