Calculating area of the loop in Folium of Descartes

Question

The curve is given as a function of a parameter $t$ as follows $$x=\frac{3at}{1+t^3}~;~y=\frac{3at^2}{1+t^3}$$ Now I am supposed to use the following formula specifically used for calculating areas enclosed by closed curves: $$\int_{t_{1}}^{t_2}\left[x(t)y'(t)-y(t)x'(t)\right]dt$$ where $t_1$ and $t_2$ are values of $t$ where the curve intersects itself. Those values should be found by setting $$\frac{3at_1}{1+t_1^3}=\frac{3at_2}{1+t_2^3}$$ and $$\frac{3at_1^2}{1+t_1^3}=\frac{3at_2^3}{1+t_2^3}$$ and then solve the system. However I am unable to do this since the system is not linear. How to proceed?

Edit: I think that this is not a duplicate question since the approach is different (no use of Green's theorem).

Answer 1

Make a drawing of the curve, and this is what you get

The curve crosses itself at $(x,y) = (0,0)$ which can only happen in two cases

$$ t= 0 ~~~\mbox{and}~~~ t\to +\infty $$

The are is then

$$ A = \int_0^{+\infty}{\rm d}t~\left(x\frac{{\rm d}y}{{\rm d}t} - y\frac{{\rm d}x}{{\rm d}t}\right) = 9a^2 \int_0^{+\infty}{\rm d}t \frac{t^2}{(1 + t^3)^3} = 3a^2 $$

Answer 2

I disagree with the previous solution. Anyone can tell from looking at the plot that the area is not equal to 3. I approached this in a different way, namely in the complex plane, where the area $A$ is given by

$$A=\frac{1}{2}\int \Im\{z^*\dot z\}dt$$

Thus,

$$z=3a\frac{t+it^2}{1+t^3}\\ z^*=3a\frac{t-it^2}{1+t^3}\\ \dot z=3a\frac{1-2t^3+i(2t-t^4)}{1+t^3}\\ \Im\{ z^*\dot z\}=9a^2\frac{t^2}{(1+t^3)^2} $$

And finally,

$$A=\frac{1}{2}\int_0^{\infty} \Im\{z^*\dot z\}dt=9a^2\frac{1}{2}\frac{1}{3}=\frac{3a^2}{2}$$

I have verified this result numerically.