The area of the region enclosed between the circles x 2 + y 2 = 4 and x 2 + ( y - 2 ) 2 = 4 is: [2026]

Question

The area of the region enclosed between the circles $x^{2} + y^{2} = 4$ and $x^{2} + {(y - 2)}^{2} = 4$ is: [2026]

Smart Achievers · Accepted Answer

(1)

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

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

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

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

$= 4 [\frac{1}{2} (\sqrt{3} + 4 \cdot \frac{π}{3}) - \sqrt{3}]$ $= 4 [\frac{2 π}{3} - \frac{\sqrt{3}}{2}]$

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

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