JEE Main 2025 Jan 22 Shift 1, Mathematics Q18: Area as Definite Integral

JEE Main2025Jan 22, Shift 1Mathematics

The area of the region, inside the circle $(x - 23)^{2} + y^{2} = 12$ and outside the parabola $y^{2} = 23 x$ is

Solution

$ y^2 = 2\sqrt{3}\,x $

$ \left(x-2\sqrt{3}\right)^2+y^2=\left(2\sqrt{3}\right)^2 $

$ A=\dfrac{\pi r^2}{2}-2\int_{0}^{2\sqrt{3}}\sqrt{2\sqrt{3}\,x}\,dx $

$ =\dfrac{\pi(12)}{2}-2\sqrt{2\sqrt{3}} \left[ \dfrac{x^{3/2}}{3/2} \right]_{0}^{2\sqrt{3}} $

$ =6\pi-16 $

Concept behind this question

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