Area (in sq.units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x^2+y^2=4 \ & \Rightarrow y^2=4-x^2\end{aligned}$ $\begin{aligned} & =\int_0^2 \sqrt{4-x^2} d x \ & =\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\cdot \frac{x}{2}ight)ight]_0^2\end{aligned}$ $=\left[\frac{2}{2} \sqrt{4-(2)^2}+\frac{4}{2} \sin ^{-1}\left(\frac{2}{2}ight)-\frac{0}{2} \sqrt{4-0}-\frac{4}{2} \sin ^{-1}\left(\frac{0}{2}ight)ight]$ $\begin{aligned} & =2 	imes \frac{\pi}{2} \ & =\pi 	ext { sq. }\end{aligned}$

Area (in sq.units) lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and…

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