The area (in sq. units) of the smaller portion enclosed between the...

To find point of intersection, eliminate $y$ from both the equation, we get

${x}^{2} + 3x - 4 = 0$

$⇒ \left(x + 4\right)\left(x - 1\right) = 0$

$x = - 4, x = 1$

Area $= \left(\mathop{∫}_{0}^{1}\sqrt{3} \cdot \sqrt{x} \cdot dx + \mathop{∫}_{1}^{2}\sqrt{4 - {x}^{2}} \cdot dx\right) \times 2$

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

$= \left(\sqrt{3}\left(\dfrac{2}{3}\right) + \left\{2 \cdot \dfrac{\pi}{2} - \left(\dfrac{\sqrt{3}}{2} + \dfrac{\pi}{3}\right)\right\}\right) \times 2$

$= \left(\dfrac{2}{\sqrt{3}} - \dfrac{\sqrt{3}}{2} + \dfrac{2\pi}{3}\right) \times 2$

$= \left(\dfrac{1}{2\sqrt{3}} + \dfrac{2\pi}{3}\right) \times 2 = \dfrac{1}{\sqrt{3}} + \dfrac{4\pi}{3}$ sq. units