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Q.
The area of the loop of the curve $ay^{2}=x^{2}(a-x)$ is
Application of Integrals
Solution:
$ay^{2}=x^{2} \left(a-x\right)$ or $y=\pm x \sqrt{\frac{a-x}{a}}$
Curve tracing: $y=x \sqrt{\frac{a-x}{a}}$
We must have $x \,\le\,a $
For $0 <\,x \le\,a, y >\,0$ and for $x <0, y <0$
Also $y=0$
$\Rightarrow x = 0, a $
Curve is symmetrical about $x$-axis
when $x \rightarrow-\infty, y \rightarrow-\infty$
Also, it can be verified that y has only one point of maxima for $0 <\,x <\,a$
Area $=2 \int\limits_{0}^{a} x \sqrt{\frac{a-x}{a}} dx \sqrt{\frac{a-x}{a}}=t$
$\Rightarrow 1-\frac{x}{a}=t^{2}$
or $x=a\left(1-t^{2}\right)$
$\Rightarrow A=2 \int\limits_{1}^{0} a \left(1-t^{2}\right)t\left(-2at\right)dt$
$=4a^{2} \int\limits_{0}^{1}\left(t^{2}-t^{4}\right)dt $
$=4a^{2}\left[\frac{t^{3}}{3}-\frac{t^{5}}{5}\right]_{0}^{1}$
$=4a^{2}\left[\frac{1}{3}-\frac{1}{5}\right]$
$=\frac{8a^{2}}{15}$ sq. units
