Question

The area of a loop of the curve $x^3+y^3=3axy$ is

• A. $a^2$
• B. $\frac{a^2}{2}$
• C. $3\frac{a^2}{2}$
• D. $5\frac{a^2}{2}$

Answer 1

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

Changing to polar form (by putting $x=r\cos \theta ,y=r\sin \theta$)
$\Rightarrow {\left(r\cos \theta \right)}^3+{\left(r\cos \theta \right)}^3=3ar\cos \theta r\sin \theta$
$\Rightarrow r^3{\cos }^3\theta +r^3{\cos }^3\theta =3a{r}^2\cos \theta \sin \theta$
$\Rightarrow r^3({\cos }^3\theta +{\sin }^3\theta )=3a{r}^2\cos \theta \sin \theta$
$\Rightarrow r=\frac{3a\cos \theta \sin \theta }{({\cos }^3\theta +{\sin }^3\theta )}$
Put $r=0,\sin \theta \cos \theta =0$
$\therefore \theta =0,\frac{\pi }{2},$ which are the limits of integration for its loop.
$\therefore$ Area of the loop$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}{r}^{2}d\theta$
$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{9a^2\sin \theta \cos \theta }{({\cos }^3\theta +{\sin }^3\theta )}d\theta$
Divide numerator and denominator by ${\cos }^6\theta$ we get
$=\frac{9a^2}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\tan }^2\theta {\sec }^2\theta }{{(1+{\tan }^3\theta )}^2}d\theta$
Put $1+{\tan }^3\theta =t$ and $3{\tan }^2\theta {\sec }^2\theta d\theta =dt$ we get
$=\frac{3a^2}{2}{\int }_1^{\infty }\frac{dt}{t^2}$
$=\frac{3a^2}{2}{\left|\frac{t^{-1}}{-1}\right|}_1^{\infty }$
$=\frac{3a^2}{2}(-0+1)$
$=\frac{3a^2}{2}$