The respective equations for the parabola and the circle are: $y^2$ = 4x ... (1) $4x^2+4y^2$ = 9 ... (2) or $x^2+y^2$ = $\left(\frac{3}{2}\right)^2$ Equation (1) is a parabola with vertex (0, 0) which opens to the right and equation (2) is a circle with centre (0, 0) and radius $\frac{3}{2}$ From equations (1) and (2), we get: $4x^2$ + 4 (4x) = 9 $4x^2$ + 16x - 9 = 0 $4x^2$ + 18x - 2x - 9 = 0 2x (2x + 9) - 1 (2x + 9) = 0 (2x + 9) (2x - 1) = 0 x = - $\frac{9}{2},\frac{1}{2}$ For x = $-\frac{9}{2}$ , $y^2$ = 4 $\left(-\frac{9}{2}\right)$ , which is not possible, hence x = $\frac{1}{2}$ Therefore, the given curves intersect at x = $\frac{1}{2}$ Required area of the region bound by the two curves = 2 $\int\limits_{0}^{\frac{1}{2}} 2\sqrt{x}dx$ + $2\int\limits_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{\frac{9}{4} - x^2}dx$ = 4 $\left|\frac{2}{3}x^{3/2}\right|_{0}^{\frac{1}{2}}$ + 2 = $\frac{8}{3}\left(\frac{1}{8}\right)^{\frac{1}{2}}$ + 2 = $\frac{8}{3}\left(\frac{1}{2\sqrt{2}}\right)$ + $\frac{9}{4}\left(\frac{\pi}{2}\right)$ - $\frac{\sqrt{2}}{2} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)$ = $\frac{2\sqrt{2}}{3} + \frac{9\pi}{8}$ - $\frac{\sqrt{2}}{2} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)$