Question

Find the area of circle $4x^2+4y^2=9$ which is interior to the parabola $x^2=4y$.

Answer 1

Given: $4x^2+4y^2=9$ ...$\left(i\right)$
$x^2=4y$
$\Rightarrow y=\frac{x^2}{4}$ ...$\left(ii\right)$
$4x^2+4y^2=9$
$\Rightarrow x^2+y^2=\frac{9}{4}={\left(\frac{3}{2}\right)}^2$
So, radius $r=\frac{3}{2}$ and centre $(0,0)$
Drawing diagram
Putting value of $x^2$ from equation $\left(ii\right)$
in equation $\left(i\right)$
$\Rightarrow 4\left(4y\right)+4y^2=9$
$\Rightarrow 4y^2+16y-9=0$
$\Rightarrow 4y^2+18y-2y-9=0$
$\Rightarrow 2y(2y+9)-1(2y+9)=0$
$\Rightarrow (2y-1)(2y+9=0)$
$\Rightarrow y=-\frac{9}{2},\frac{1}{2}$

As $y=\frac{x^2}{4}$, so it is always positive,
$\therefore y=\frac{1}{2}$

$\Rightarrow \frac{x^2}{4}=\frac{1}{2}$
$\Rightarrow x^2=2$
$\Rightarrow x=\pm \sqrt{2}$

Hence, the point of intersections are
$A=(-\sqrt{2},\frac{1}{2})$ & $C=(\sqrt{2},\frac{1}{2})$

Area required
$=2\times AreaBOCB$

$=2[{\int }_0^{\sqrt{2}}{y}_{1}dx-{\int }_0^{\sqrt{2}}{y}_{2}dx]$

Where $y_1=\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}$ & $y_2=\frac{x^2}{4}$

$=2[{\int }_0^{\sqrt{2}}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}dx-{\int }_0^{\sqrt{2}}\frac{x^2}{4}dx]$
$=2[{\int }_0^{\sqrt{2}}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}dx-\frac{1}{4}{\int }_0^{\sqrt{2}}{x}^{2}dx]$

$=2\left[\begin{array}{c}{[\frac{x}{2}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}+\frac{{\left(\frac{3}{2}\right)}^2}{2}{\sin }^{-1}\frac{x}{\frac{3}{2}}]}_0^{\sqrt{2}}\\ -\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^{\sqrt{2}}\end{array}\right]$

$=2[[\frac{1}{\sqrt{2}}\sqrt{\frac{1}{4}}+\frac{9}{8}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)]-\left[\frac{\sqrt{2}}{6}\right]]$

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

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

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

Therefore, required area is

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