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.