Question

The area of the loop of the folium $x^3+y^3-3axy=0$ is given by

• A. $\frac{1}{2}{a}^2$
• B. $a^2$
• C. $\frac{3}{2}{a}^2$
• D. $2a^2$

Answer 1

The correct option is C $\frac{3}{2}{a}^2$

Changing the equation of the curve $x^3+y^3-3axy=0$ into polar form by substituting
$x=r\cos \theta$ and $y=r\sin \theta$
We have, ${\left(r\cos \theta \right)}^3+{\left(r\sin \theta \right)}^3-3a(r\cos \theta \times r\sin \theta )=0$
Or $r=\frac{\left(3a\cos \theta \sin \theta \right)}{\left({\cos }^3\theta \right)+\left({\sin }^3\theta \right)}\cdots \cdots \left(1\right)$
From (1) $r=0$ when $\theta =0$ and $\theta =\frac{\pi }{2}$
Hence the required area of the loop is,
$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{r}^{2}d\theta =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{\left(\frac{3a\cos \theta \sin \theta }{\left({\cos }^3\theta \right)+\left({\sin }^3\theta \right)}\right)}^{2}d\theta$
$=\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}{\left(\frac{\cos \theta \sin \theta }{\left({\cos }^3\theta \right)+\left({\sin }^3\theta \right)}\right)}^{2}d\theta$
$=\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\left(\frac{{\tan }^2{\theta \sec }^2\theta }{{(1+{\tan }^3\theta )}^2}\right)d\theta$
Dividing numerator and denominator by ${\cos }^6\theta$
Now substitute $(1+{\tan }^3\theta )=t$
$\Rightarrow 3{\tan }^2{\theta \sec }^2\theta d\theta =dt$
Also when $\theta =0,t=1$
and when $\theta \to \frac{\pi }{2},t\to \infty$
$\therefore$ Area of the loop $=\frac{3a^2}{2}{\int }_1^{\infty }\frac{1}{t^2}dt$
$=\frac{3a^2}{2}{[-\frac{1}{t}]}_1^{\infty }=\frac{3a^2}{2}$