Find the area of curve The folium of descartes is defined by
$$x^3+y^3-3xy=0$$
(cartesian)
$$r=\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta}$$(polar) or
$$x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}$$(parametric).
It looks like this and has the slant asymptote $y=-x-1$. I'm trying to show that the area between the asymptote and the infinite branches of the curve is the same as the area of the loop. I already know the area of the loop, it is $1.5$ units.
I managed to show that the area between the asymptote and the branches is also $1.5$ units, but the method I used was last-resort inelegant** (I, or more specifically wolfram-alpha, solved the cartesian equation for $y$ and then the resultant integral).
Are there different ways to show that the asymptote-branches area is 1.5 units? Showing that its area is the same as the area of the loop is fine as well, but probably harder/needing a special trick.
Thank you!
** The problem source says about this problem: "Use a CAS to evaluate the resultant integral" so I didn't feel too guilty about making WA do it. The problem source is James Stewart Calculus, in the polar and parametric equations chapter.
 A: Introduce new coordinates $\xi$, $\eta$ such that the asymptote becomes the $\xi$-axis:
$$\xi={1\over\sqrt{2}}(x-y),\quad \eta={1\over\sqrt{2}}(x+y+1)\ .$$
(This amounts to a $45^\circ$ rotation and a shift in the $\eta$ direction.) The parametric representation of the folium then is
$$\xi(t)={1\over\sqrt{2}}{3(t-t^2)\over1+t^3},\quad\eta={1\over\sqrt{2}}\left({3(t+t^2)\over1+t^3}+1\right)\ .\tag{1}$$
The branch $\gamma$ of the folium going from $\bigl(0,{1\over\sqrt{2}}\bigr)$ to $(+\infty,0)$  can now be viewed as  a graph over the $\xi$-axis, albeit not parametrized by $\xi$. Instead it is parametrized by $(1)$ with the $t$-interval $-\infty<t<-1$. In order to find the area between $\gamma$ and the $\xi$-axis we have to compute 
$$\int_\gamma \eta\>d\xi=\int_{-\infty}^{-1}\eta(t)\,\xi'(t)\>dt=\int_{-\infty}^{-1}{3(1-2t-2t^3+t^4)\over2(1-t+t^2)^3}\>dt={3\over4}\ ,$$
which is half the desired area. (I have used Mathematica to obtain the final value.)
A: Great answer from Christian Blatter in 2018.
If you want to explicitly solve the integral:
$$\int_{}^{}{1-2t-2t^3+t^4\over(1-t+t^2)^3}\>dt,$$
the anti-derivative is
$$1-t+\frac{3}{2}t^2-t^3\over(1-t+t^2)^2$$
which tends to 0 at +/- $\infty$, and equals $\frac{1}{2}$ at +/-1.
A: Integrate the area in polar coordinates
\begin{align}
Area&=\frac12\int_0^{\pi/2} r^2(\theta)d\theta \\
&= \frac12\int_0^{\pi/2}\left(\frac{3\sec \theta \tan \theta}{1+\tan^3 \theta}\right)^2 d\theta=\frac12\left(-\frac3{1+\tan^3\theta}\right)_0^{\pi/2}=\frac32
\end{align}
