$\text{ The area bounded by the loop of }4y^2=x^2\left(4-x^2\right)\text{ is }$

$\begin{array}{l}4y^2=x^2\left(4-x^2\right)---\left(i\right)\\ \Rightarrow y=\frac{1}{2}\sqrt{x^2\left(4-x^2\right)}\Rightarrow y=\frac{x}{2}\sqrt{\left(4-x^2\right)}\\ \therefore \text{ Area }\left(A\right)=4\times {\int }_0^2 \frac{2x}{2\times 2}\sqrt{\left(4-x^2\right)}dx\\ \text{ Put }\left(4-x^2\right)=t\Rightarrow -2xdx=dt\\ \therefore A=\frac{-4}{4}{\int }_4^0 \sqrt{t}dt={\int }_0^4 \sqrt{t}dt={\left[\frac{t^{3/2}}{3/2}\right]}_0^4=\frac{2}{3}\times [\sqrt{64}-0]\\ \Rightarrow A=\frac{16}{3}\text{ sq.units }\end{array}$