Question

Find the area of the region, {$(x,y):y^2\le 4x,4x^2+4y^2\le 9$} using method of integration.

Answer 1

The point of intersection of both the curves are $(\frac{1}{2},\sqrt{2})$ and $(\frac{1}{2},-\sqrt{2})$

The required area is given by $OABCO$.

It can be observed that area $OABCO$ is symmetrical about $x-$axis.

Therefore, Area $OABCO=2\times$ Area of $OBC$

Area $OBCO=$ Area $OMC+$ Area $MBC$

$={\int }_0^{\frac{1}{2}}2\sqrt{x}dx+={\int }_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{2}\sqrt{9-4x^2}dx$

$={\int }_0^{\frac{1}{2}}2\sqrt{x}dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{2}\sqrt{3^2-(2x{)}^2}dx$

$=2{\left[2x^{\frac{3}{2}}\right]}_0^{\frac{1}{2}}+\frac{1}{2}{\left[\frac{9}{4}{\sin }^{-1}(\frac{2x}{3}+\frac{1}{2}\sin \left(2{\sin }^{-1}\left(\frac{2x}{3}\right)\right))\right]}_{\frac{1}{2}}^{\frac{3}{2}}+c$

$=\frac{2}{3\sqrt{2}}+\frac{9}{4}{\sec }^{-1}\left(3\right)-\frac{1}{\sqrt{2}}$