Question

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

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

Answer 1

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

$4y^2=x^2(4-x^2)$

$\Rightarrow y=\pm \frac{1}{2}\sqrt{x^2(4-x^2)}$

$\Rightarrow y=\pm \frac{x}{2}\sqrt{(4-x^2)}$

Therefore, $Area\left(A\right)=4\times {\int }_0^2\frac{x}{2}\sqrt{4-x^2}dx$

Let $4-x^2=t\Rightarrow -2xdx=dt$

$\Rightarrow A={\int }_0^4\sqrt{t}dt=[\frac{t^{3/2}}{3/2}{]}_0^4$

$\Rightarrow A=\frac{2}{3}\sqrt{64}$

$\Rightarrow A=\frac{16}{3}$sq.units