Question

The area enclosed by the curve $a^2{x}^2=y^3(2a-y)$ is:

• A. $\pi a^2$
• B. $\pi a^3$
• C. $\pi a$
• D. $\pi a^4$

Answer 1

The correct option is A $\pi a^2$

Let us find the Limits of integration.
$\left(i\right)$The curve is symmetrical about $y-$axis.
$\left(ii\right)$It passes through the origin and the tangents at the origin are $x^2=0$ or $x=0,x=0$
$\therefore$ There is a cusp at the origin.
$\left(iii\right)$The curve has no asymptote.
$\left(iv\right)$The curve meets the $x-$axis at the origin only and meets the $y-$axis at $(0,2a)$.From the equation of the curve, we have
$x=\frac{y}{a}\sqrt{y(2a-y)}$
For $y<0$ or $y>2a,x$ is imaginary.
Thus the curve entirely lies between $y=0,x-$axis and $y=2a,$ which is shown in the figure.
$\therefore$ Area of the curve$=2{\int }_0^{2a}x.dy$
Put $y=2a{\sin }^2\theta \therefore dy=4a\sin \theta \cos \theta d\theta$
$\Rightarrow =\frac{2}{a}{\int }_0^{2a}y\sqrt{\left[y(2a-y)\right]}$
$=\frac{2}{a}{\int }_0^{\frac{\pi }{2}}2a{\sin }^2\theta \sqrt{2a{\sin }^2\theta (2a-2a{\sin }^2\theta )}\times 4a\sin \theta \cos \theta d\theta$
$=32a^2{\int }_0^{\frac{\pi }{2}}{\sin }^4\theta {\cos }^2\theta d\theta$
$=32a^2\frac{3.1\times 1}{\mathrm{6.4.2}}.\frac{\pi }{2}=\pi a^2$