Mathematics · Area under the Curves

JEE Main 2026 — 23 January, Evening Shift — Question 12

The area of the region enclosed between the circles $x^{2}+y^{2}=4$ and $x^{2}+(y-2)^{2}=4$ is :

Answer: D

$A=2 \int_{0}^{\sqrt{3}}\left[\sqrt{4-x^{2}}-\left(2-\sqrt{4-x^{2}}\right)\right] d x$

$=2 \int_{0}^{\sqrt{3}}\left(2 \sqrt{4-x^{2}}-2\right) d x$ $=4 \int_{0}^{\sqrt{3}}\left(\sqrt{4-x^{2}}-1\right) d x$

$=\left[4\left[\frac{1}{2}\left(x \sqrt{4-x^{2}}+4 \sin ^{-1} \frac{x}{2}\right)-x\right]\right]_{0}^{\sqrt{3}}$

$=4\left[\frac{1}{2}\left(\sqrt{3}+4 \times \frac{\pi}{3}\right)-\sqrt{3}\right]=4\left[\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right]$

$=\frac{8 \pi}{3}-2 \sqrt{3}$ (Sq. units)

Answer key and solution verified before publishing.

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