Find the area of the circle
Question:

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

Solution:

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle, $4 x^{2}+4 y^{2}=9$, and parabola, $x^{2}=4 y$, we obtain the point of intersection as $\mathrm{B}\left(\sqrt{2}, \frac{1}{2}\right)$ and $\mathrm{D}\left(-\sqrt{2}, \frac{1}{2}\right)$.

It can be observed that the required area is symmetrical about $y$-axis.

$\therefore$ Area $\mathrm{OBCDO}=2 \times$ Area $\mathrm{OBCO}$

We draw BM perpendicular to $\mathrm{OA}$.

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

Therefore, Area $\mathrm{OBCO}=$ Area $\mathrm{OMBCO}-$ Area $\mathrm{OMBO}$

$=\int_{0}^{\sqrt{2}} \sqrt{\frac{\left(9-4 x^{2}\right)}{4}} \mathrm{~d} x-\int_{0}^{\sqrt{2}} \frac{x^{2}}{4} \mathrm{~d} x$

$=\int_{0}^{\sqrt{2}} \sqrt{\left(\frac{3}{2}\right)^{2}-x^{2}} \mathrm{~d} x-\frac{1}{4} \int_{0}^{\sqrt{2}} x^{2} \mathrm{~d} x$

$=\left.\left(\frac{x}{2} \sqrt{\left(\frac{3}{2}\right)^{2}-x^{2}}+\frac{9}{8} \sin ^{-1} \frac{2 x}{3}\right)\right|_{0} ^{\sqrt{2}}-\left.\frac{1}{4}\left(\frac{x^{3}}{3}\right)\right|_{0} ^{\sqrt{2}}$

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

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

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

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

Therefore, the required area $\mathrm{OBCDO}$ is $\left(2 \times \frac{1}{2}\left[\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right]\right)=\left[\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\right]$ units

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