The area (in sq units) enclosed by the loop of the curve $a y^2=x^2(a-x),(a0)$ is

Question

The area (in sq units) enclosed by the loop of the curve $a y^2=x^2(a-x),(a>0)$ is

Aicharya · Accepted Answer

Given curve is $a y^2=x^2(a-x),(a > 0)$ The required area $\begin{aligned} & =2 \int_0^a x \sqrt{\frac{a-x}{a}} d x \ & =\frac{2}{\sqrt{a}} \int_0^a x \sqrt{a-x} d x \end{aligned}$ Put $a-x=t^2$, at $x=0, t=\sqrt{a}$ and , at $x=a, t=0$ and $-d x=2 t d t$ So, required area $=-\int_{\sqrt{a}}^0 \frac{4}{\sqrt{a}} \times t^2\left(a-t^2\right) d t$ $\begin{aligned} & =\frac{4}{\sqrt{a}} \int_0^{\sqrt{a}}\left(a t^2-t^4\right) d t=\frac{4}{\sqrt{a}}\left[\frac{a t^3}{3}-\frac{t^5}{5}\right]_0^{\sqrt{a}} \ & =\frac{4}{\sqrt{a}}\left(\frac{a^2 \sqrt{a}}{3}-\frac{a^2 \sqrt{a}}{5}\right)=\frac{8}{15} a^2 \end{aligned}$ Hence, option (d) is correct.

The area (in sq units) enclosed by the loop of the curve \(a y^2=x^2(a-x),(a>0)\) is

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