Question

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

Answer 1

Given, circle is $4x^2+4y^2=9$ and the given parabola is $x^2=4y.$ The two curves meet where

$4\left(4y\right)+4y^2=9\Rightarrow 4y^2+16y-9=0\Rightarrow y=\frac{-16\pm \sqrt{256+144}}{2\times 4}=\frac{-16\pm \sqrt{400}}{8}=\frac{-16\pm 20}{8}=\frac{1}{2},-\frac{9}{2}$
But y > 0, therefore the two curves meet when $y=\frac{1}{2}.$
i.e., when $x^2=4\times \frac{1}{2}=2$, i.e., when $x=\pm \sqrt{2}.$
$\therefore$ Required area (shown in shaded region)
$=2{\int }_0^{\sqrt{2}}(y_2-y_1)dx=2\{{\int }_0^{\sqrt{2}}\sqrt{\frac{9-4x^2}{4}}dx-{\int }_0^{\sqrt{2}}\frac{x^2}{4}dx\}=2{\int }_0^{\sqrt{2}}\sqrt{(\frac{3}{2}{)}^2-x^2}dx-\frac{2}{4}[\frac{x^3}{3}{]}_0^{\sqrt{2}}$

$=2[\frac{x}{2}\sqrt{(\frac{3}{2}{)}^2-x^2}+\frac{(\frac{3}{2}{)}^2}{2}si{n}^{-1}(\frac{x}{\frac{3}{2}}){]}_0^{\sqrt{2}}-\frac{1}{6}[(\sqrt{2}{)}^3-0]=[x\sqrt{\frac{9}{4}-x^2}+\frac{9}{4}si{n}^{-1}(\frac{2x}{3}){]}_0^{\sqrt{2}}-\frac{2\sqrt{2}}{6}=[\sqrt{2}\sqrt{\frac{9}{4}-2}+\frac{9}{4}si{n}^{-1}\left(\frac{2\sqrt{2}}{3}\right)]-(0-0)-\frac{2\sqrt{2}}{6}=\frac{\sqrt{2}}{2}+\frac{9}{4}si{n}^{-1}(\frac{2\sqrt{2}}{3})-\frac{2\sqrt{2}}{6}=(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{3})+\frac{9}{4}si{n}^{-1}(\frac{2\sqrt{2}}{3})=\frac{\sqrt{2}}{6}+\frac{9}{4}si{n}^{-1}(\frac{2\sqrt{2}}{3})$ sq unit