Using integration, find the area of the region bounded by the curves $x^2 + y^2 = 4$, $x = \sqrt{3}y$ and X-axis lying in the first quadrant.

$\frac{\pi}{3}$ sq units

The correct answer is Option (1) → $\frac{\pi}{3}$ sq units

Given equation of circle:

$x^2 + y^2 = 4$

or $x^2 + y^2 = (2)^2$

$∴\text{radius} = 2$

So, point $A$ is $(2, 0)$ and point $B$ is $(0, 2)$. Let line $x = \sqrt{3}y$ intersect the circle at point $C$.

On solving $x^2 + y^2 = 4$ and $x = \sqrt{3}y$, we get:

$(\sqrt{3}y)^2 + y^2 = 4$

$⇒3y^2 + y^2 = 4$

$⇒4y^2 = 4 ⇒y^2 = 1 ⇒y = \pm 1$

For $y = 1, x = \sqrt{3}$ and for $y = -1, x = -\sqrt{3}$.

Since point $C$ is in the $1^{\text{st}}$ quadrant, $C = (\sqrt{3}, 1)$.

$∴\text{Required area} = \int\limits_{0}^{\sqrt{3}} y_{\text{line}} dx + \int\limits_{\sqrt{3}}^{2} y_{\text{circle}} dx$

$= \int\limits_{0}^{\sqrt{3}} \frac{x}{\sqrt{3}} dx + \int\limits_{\sqrt{3}}^{2} \sqrt{4 - x^2} dx$

$= \frac{1}{\sqrt{3}}\int\limits_{0}^{\sqrt{3}} x\, dx + \int\limits_{\sqrt{3}}^{2} \sqrt{4 - x^2} dx$

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

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

$= \frac{\sqrt{3}}{2} + 2\sin^{-1}(1) - \frac{\sqrt{3}}{2} - 2\sin^{-1}\frac{\sqrt{3}}{2}$

$= 2 \cdot \frac{\pi}{2} - 2 \cdot \frac{\pi}{3} = \pi - \frac{2\pi}{3} = \mathbf{\frac{\pi}{3} \text{ sq. units}}$