Question

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

Answer 1

The required area is represented by the shaded area $OBCDO$.
Solving the given equation of circle, $4x^2+4y^2=9$, and parabola, $x^2=4y$, we obtain the point of intersection as $B(\sqrt{2},\frac{1}{2})$ and $D(-\sqrt{2},\frac{1}{2})$.
It can be observed that the required area is symmetrical about $y$-axis.

$\therefore$ Area $OBCDO$ $=2\times AreaOBCO$

We draw $BM$ perpendicular to $OA$.

Therefore, the coordinates of M are $(\sqrt{2},0)$.

Therefore, Area $OBCO$ $=$ Area $OMBCO$ - Area $OMBO$

$={\int }_0^{\sqrt{2}}\sqrt{\frac{(9-4x^2)}{4}}dx-{\int }_0^{\sqrt{2}}\frac{x^2}{4}dx$

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

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

$=\frac{1}{4}[\sqrt{2}\sqrt{9-8}+\frac{9}{2}{\sin }^{-1}\frac{2\sqrt{2}}{3}]-\frac{1}{12}(\sqrt{2}{)}^3$

$=\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}$

$=\frac{\sqrt{2}}{12}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}$

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

Therefore, the required area $OBCDO$ is

$(2\times \frac{1}{2}[\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}])=[\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}]$ sq. units