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Area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$, the line $x=\sqrt{3} y$ and $x$-axis is
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$ sq units
Required area
$=\int_{0}^{1}\left(x_{2}-x_{1}\right) d y$
$=\int_{0}^{1}\left(\sqrt{4-y^{2}}-\sqrt{3} y\right) d y$
$=\left[\frac{1}{2} y \sqrt{4-y^{2}}+\frac{1}{2}(4) \sin ^{-1} \frac{y}{2}-\frac{\sqrt{3} y^{2}}{2}\right]_{0}^{1}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1} 0 \\
=\frac{\sqrt{3}}{2}+2\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}=\frac{\pi}{3} \text { sq units }
\end{array}
$$
$=\int_{0}^{1}\left(x_{2}-x_{1}\right) d y$
$=\int_{0}^{1}\left(\sqrt{4-y^{2}}-\sqrt{3} y\right) d y$
$=\left[\frac{1}{2} y \sqrt{4-y^{2}}+\frac{1}{2}(4) \sin ^{-1} \frac{y}{2}-\frac{\sqrt{3} y^{2}}{2}\right]_{0}^{1}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{2}+2 \sin ^{-1}\left(\frac{1}{2}\right)-\frac{\sqrt{3}}{2}-2 \sin ^{-1} 0 \\
=\frac{\sqrt{3}}{2}+2\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}=\frac{\pi}{3} \text { sq units }
\end{array}
$$
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