Question

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

• A. $2\pi a^2$
• B. $\frac{\pi }{3}{a}^2$
• C. $\frac{4}{15}{a}^2$
• D. $\frac{8}{15}{a}^2$

Answer 1

Answer: (D)

Given curve is
$a{y}^2=x^2(a-x),(a>0)$

The required area
$=2\underset{0}{\overset{a}{\int }}x\sqrt{\frac{a-x}{a}}dx$
$=\frac{2}{\sqrt{a}}{\int }_0^{a}x\sqrt{a-x}dx$
Put $a-x=t^2$, at $x=0,t=\sqrt{a}$ and,
at $x=a,t=0$
and $-dx=2tdt$
So, required area $=-\underset{\sqrt{a}}{\overset{0}{\int }}\frac{4}{\sqrt{a}}\times t^2\left(a-t^2\right)dt$
$=\frac{4}{\sqrt{a}}{\int }_0^{\sqrt{a}}\left(a{t}^2-t^4\right)dt=\frac{4}{\sqrt{a}}{\left[\frac{a{t}^3}{3}-\frac{t^5}{5}\right]}_0^{\sqrt{a}}$
$=\frac{4}{\sqrt{a}}\left(\frac{a^2\sqrt{a}}{3}-\frac{a^2\sqrt{a}}{5}\right)=\frac{8}{15}{a}^2$