Question

The area bounded by the loop of the curve $4y^2=x^2(4-x^2)$ is

• A. $\frac{7}{3}$ sq. units
• B. $\frac{8}{3}$ sq. units
• C. $\frac{11}{3}$ sq. units
• D. $\frac{16}{3}$ sq. units

Answer 1

The correct option is D $\frac{16}{3}$ sq. units

$4y^2=x^2(4-x^2)$
$\Rightarrow y=\pm \frac{1}{2}\sqrt{x^2(4-x^2)}$
$\Rightarrow y=\pm \frac{x}{2}\sqrt{(4-x^2)}$
$\therefore$ Area $\left(A\right)=4\times {\int }_0^2\frac{x}{2}\sqrt{(4-x^2)}dx$
Let $4-x^2=t\Rightarrow -2xdx=dt$
$\Rightarrow A={\int }_0^4\sqrt{t}dt={\left[\frac{t^{3/2}}{3/2}\right]}_0^4=\frac{2}{3}\times [\sqrt{64}-0]$
$\Rightarrow A=\frac{16}{3}$ sq. units.