CBSE Class 12 · 2007 · Area Under CurvesLong Answer

The circle is $x^2+y^2=\frac{9}{4}$ (centre $O$, radius $\frac{3}{2}$) and the parabola is $y^2=4x$.
Intersection: substitute $x=\frac{y^2}{4}$ into the circle, $\frac{y^4}{16}+y^2=\frac{9}{4}$. With $t=y^2$, $t^2+16t-36=0\Rightarrow (t-2)(t+18)=0$, so $t=2$, i.e. $y=\pm \sqrt{2}$ (and $x=\frac{1}{2}$).
Integrating in $y$ (right boundary = circle, left boundary = parabola), and using symmetry about the $x$-axis:

Now $2{\int }_0^{\sqrt{2}}\sqrt{\frac{9}{4}-y^2}dy=[y\sqrt{\frac{9}{4}-y^2}+\frac{9}{4}{\sin }^{-1}\frac{2y}{3}{]}_0^{\sqrt{2}}=\frac{\sqrt{2}}{2}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3},$ and $2{\int }_0^{\sqrt{2}}\frac{y^2}{4}dy=\frac{\sqrt{2}}{3}.$

Since ${\sin }^{-1}\frac{2\sqrt{2}}{3}=\frac{\pi }{2}-{\sin }^{-1}\frac{1}{3}$, this equals $\frac{\sqrt{2}}{6}+\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}$.