Question

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

Answer 1

Let the given equations

Equation of the circle is

$4x^2+4y^2=9$ ------ $\left(i\right)$

And equation of parabola is

$x^2=4y$

$y=\frac{x^2}{4}$ ---- $\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$\begin{array}{c}4x^2+4{\left(\frac{x^2}{4}\right)}^2=9\\ \\ x^4+16x^2-36=0\\ \\ \left(x^2+18\right)\left(x^2-2\right)=0\\ \\ x=\pm \sqrt{2}\ne \sqrt{-18}\end{array}$

Required area

$\begin{array}{c}=2{\int }_0^{\sqrt{2}}\left(y_1-y_2\right)dx\\ \\ =2\left[{\int }_0^{\sqrt{2}}\left\{\sqrt{\frac{9}{4}-x^2}-\frac{x^2}{4}\right\}dx\right]\left(y_1:x^2+y^2,y_2:x^2=4y\right)\\ \\ =2{\left[\frac{x}{2}\sqrt{\frac{9}{4}-x^2}+\frac{9}{8}{\sin }^{-1}\frac{x}{\frac{3}{2}}-\frac{x^3}{12}\right]}_0^{\sqrt{2}}\\ \\ =2\left[\frac{\sqrt{2}}{4}+\frac{9}{8}{\sin }^{-1}\frac{2\sqrt{2}}{3}-\frac{\sqrt{2}}{6}\right]=\left(\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}\right)sq.units\end{array}$