Aspire Faculty ID #18022 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

The area of the region enclosed between the circles

$x^2 + y^2 = 4$ and $x^2 + (y-2)^2 = 4$ is:

Solution



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

$= 2\int_{0}^{\sqrt{3}} (2\sqrt{4-x^2} - 2),dx$

$= 4\int_{0}^{\sqrt{3}} (\sqrt{4-x^2} - 1),dx$

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

$= \frac{8\pi}{3} - 2\sqrt{3}$

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