The area of the region, inside the circle (x-2 √(3)) 2 +y 2 =12 and outside the parabola y 2 =2 √(3) x is

Mathematics · Area under the Curves

JEE Main 2025 — 22 January, Morning Shift — Question 18

The area of the region, inside the circle $(x-2 \sqrt{3})^{2}+y^{2}=12$ and outside the parabola $y^{2}=2 \sqrt{3} x$ is

\mathrm{Equation\ of\ circle:}\quad (x-2\sqrt{3})^{2}+y^{2}=12 \quad \mathrm{(center}\ (2\sqrt{3},0), \, \mathrm{radius}\ 2\sqrt{3}\mathrm{).}

\mathrm{Equation\ of\ parabola:}\quad y^{2}=2\sqrt{3}\,x \quad \Rightarrow \quad x=\frac{y^{2}}{2\sqrt{3}}.

\mathrm{Intersection\ points:\ substitute}\ x=\frac{y^{2}}{2\sqrt{3}} \ \mathrm{into\ the\ circle:}

\left(\frac{y^{2}}{2\sqrt{3}}-2\sqrt{3}\right)^{2}+y^{2}=12.

\frac{y^{2}(y^{2}-12)}{12}=0 \quad \Rightarrow \quad y=0, \, y=\pm 2\sqrt{3}.

\mathrm{Thus,\ relevant\ interval:}\quad y\in[-2\sqrt{3},2\sqrt{3}].

\mathrm{For\ a\ given\ } y:\quad x_{\mathrm{left\ circle}}=2\sqrt{3}-\sqrt{12-y^{2}}, \qquad x_{\mathrm{parabola}}=\frac{y^{2}}{2\sqrt{3}}.

\mathrm{So\ required\ area:}\quad A=\int_{-2\sqrt{3}}^{2\sqrt{3}} \left(\frac{y^{2}}{2\sqrt{3}}-\Big(2\sqrt{3}-\sqrt{12-y^{2}}\Big)\right) dy.

\mathrm{By\ symmetry:}\quad A=2\int_{0}^{2\sqrt{3}} \left(\frac{y^{2}}{2\sqrt{3}}-2\sqrt{3}+\sqrt{12-y^{2}}\right) dy.

\frac{2}{2\sqrt{3}}\int_{0}^{2\sqrt{3}}y^{2}\,dy = \frac{1}{\sqrt{3}}\cdot \frac{(2\sqrt{3})^{3}}{3} = 8,

-2\cdot 2\sqrt{3}\cdot 2\sqrt{3} = -24,

2\int_{0}^{2\sqrt{3}}\sqrt{12-y^{2}}\,dy = 2 \cdot \frac{\pi (2\sqrt{3})^{2}}{4} = 6\pi.

\mathrm{Therefore:}\quad A = 8 - 24 + 6\pi = 6\pi - 16.

\boxed{6\pi - 16}

Answer key and solution verified before publishing.

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