# Using integration, find the area of the region bounded by the parabola $y^{2}= 4x$ and the circle $4x^{2}+4y^{2}= 9.$

for the circle $4x^{2}+4y^{2}= 9,\; we\; have$
$x^{2}+y^{2}= \left ( \frac{3}{2} \right )^{2}$
$\therefore centre\; is\; at\left ( 0,0 \right )\S \; r= \frac{3}{2}$
Also vertex of the parabola $y^{2}= 4x\; is\; at\left ( 0,0 \right ).$ solving the equations of curves, we get:
$4x^{2}+16x-9= 0\Rightarrow \left ( 2x-1 \right )\left ( 2x+9 \right )= 0$
$\therefore x= \frac{1}{2}\left [ x\not\equiv \frac{-9}{2} \right ]$
As the circle and parabola, both are symmetrical about the x-axis
$\therefore Required\; area= 2\left [ \int_{0}^{\frac{1}{2}} 2\sqrt{x}\, dx +\int_{\frac{1}{2}}^{\frac{3}{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}dx}\right ]$
$\Rightarrow 2\left [ \frac{4}{3}\left [ x\sqrt{x} \right ] \right ]^{\frac{1}{2}}_{0}+\left [ \frac{x}{2}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}+\frac{9}{8}\sin^{-1}\frac{2x}{3}\right ]^{\frac{3}{2}}_{\frac{1}{2}}$
$\Rightarrow 2\left [ \frac{4}{3} \left [ \frac{1}{2\sqrt{2}}-0 \right ]+\left [ \frac{3}{4} \sqrt{0}+\frac{9}{8}\sin^{-1}1\right ]-\left [ \frac{1}{4}\frac{\sqrt{9-1}}{\sqrt{4}} +\frac{9}{8}\sin^{-1}\frac{1}{3}\right ]\right ]$
$\Rightarrow 2\left [ \frac{2\sqrt{2}}{6}+\frac{9}{8} \times \frac{\pi }{2}-\frac{\sqrt{2}}{4}-\frac{9}{8}\sin^{-1}\frac{1}{3}\right ]$
$\Rightarrow \left [ \frac{\sqrt{2}}{6}+\frac{9}{4} \cos^{-1}\frac{1}{3}\right ]sq\cdot units$

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