Question

Find the area of the circle 4x2+4y2=9 which is interior to the parabola x2=4y

Answer 1

The required area is represented by the shaded area OBCDO.

Solving the given equation of circle,4x2+4y2=9,and parabola,x2=4y,we obtain the

point of intersection as B$(\sqrt2,\frac{1}{2})$and D $(-\sqrt2,\frac{1}{2}).$

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

∴Area OBCDO=2×Area OBCO

We draw BM perpendicular to OA.

Therefore,the coordinates of M are $(\sqrt2,0)$.

Therefore,Area OBCO=Area OMBCO-Area OMBO

=$∫_0^{\sqrt2}\sqrt{\frac{(9-4x^2)}{4}}dx-∫_0^{\sqrt2}\sqrt{\frac{x^2}{4}}dx$

=$\frac{1}{2}∫_0^{√2}\sqrt{9-4x^2}\,dx-\frac{1}{4}∫_0^{\sqrt2}x^2dx$

=$\frac{1}{4}\bigg[x\sqrt{9-4x^2}+\frac{9}{2}sin^{-1}\frac{2x}{3}\bigg]_0^{\sqrt2}-\frac{1}{4}\bigg[\frac{x^3}{3}\bigg]_0^{\sqrt2}$

=$\frac{1}{4}[\sqrt{2}\sqrt{9-8}+\frac{9}{2}sin^{-1}\frac{2√2}{3}]-\frac{1}{12}(\sqrt2)^3$

=$\frac{\sqrt2}{4}+\frac{9}{8}sin^{-1}\frac{2\sqrt2}{3}-\frac{\sqrt2}{6}$

=$\frac{\sqrt2}{12}+\frac{9}{8}sin^{-1}\frac{2\sqrt2}{3}$

=$\frac{1}{2}(\frac{\sqrt2}{6}+\frac{9}{4}sin^{-1}\frac{2\sqrt2}{3})$

Therefore,the required area OBCDO is

$(2×\frac{1}{2}[\frac{\sqrt2}{6}+\frac{9}{4}sin^{-1}\frac{2\sqrt2}{3}])=[\frac{\sqrt2}{6}+\frac{9}{4}sin^{-1}\frac{2\sqrt2}{3}]$units.