Question

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

Answer 1

Given curves are $y^2=4x$ ....(1)

$4x^2+4y^2=9$ ...(2)

$\Rightarrow x^2+y^2=\frac{9}{4}$

Center of circle is (0,0) and radius of circle is $\frac{3}{2}$

Put the value from eqn (1) in eqn (2),

$4x^2+16x-9=0$

$\Rightarrow x=\frac{1}{2},-\frac{9}{2}$

But $x=-\frac{9}{2}$, not possible .

So, $x=\frac{1}{2}$

$\Rightarrow y=\pm \sqrt{2}$

So, the points of intersection of both the curves are $(\frac{1}{2},\sqrt{2})$ and $(\frac{1}{2},-\sqrt{2})$.

The shaded region $OABCO$ represents the area bounded by the curves $\left\{\right(x,y):y^2\le 4x,4x^2+4y^2\le 9\}$

Since, the area $OABCO$ is symmetrical about x-axis.
$\therefore AreaOABCO=2\times AreaOBC$

$AreaOBCO=AreaOMC+AreaMBC$

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

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

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

$=\frac{\sqrt{2}}{3}+[0+\frac{9}{8}{\sin }^{-1}\left(1\right)-\frac{\frac{1}{2}\sqrt{2}}{2}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)]$

$=\frac{\sqrt{2}}{3}+\frac{9\pi }{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)$

Required Area $=2\left[\frac{\sqrt{2}}{3}+\frac{9\pi }{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)\right]$

$=\frac{\sqrt{2}}{6}+\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\left(\frac{1}{3}\right)$ sq.units